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There is an object in the universe for which the Sun appears to be more stationary as it travels through the universe than a star on some other system. There is probably another object for which both our sun and the 2nd star (combined) appear to be more stationary. Does there exist in an object in the universe for which all objects appear to be the most stationary (on average)? What about an object that makes all other objects appear to be the least stationary? Would identifying this object have any significance at all or is it just a random rock somewhere in space that has some coordinates that will fail to offer additional insight? Wouldn't this object experience the least amount of time dilation (i.e. everything else is in slow motion with respect to this object)?

Qmechanic
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    This sounds like Kant's idea of attaching a reference frame to the "center of mass of the universe". Inductively we first take the center of mass of the solar system, then of the galaxy, etc. This was supposed to give physical meaning to Newton's absolute space, the ultimate inertial frame. https://books.google.com/books?id=5rxYBvx7tW0C&pg=PA69&lpg=PA69&dq=kant+frame+center+of+mass&source=bl&ots=LCtYVh8v-U&sig=LEF6XJXCV4L1ejc2j5nkoRr9Uzs&hl=en&sa=X&ved=0ahUKEwjttdW00qrJAhVHnIgKHa5AAMoQ6AEIJDAD#v=onepage&q=kant%20frame%20center%20of%20mass&f=false – Conifold Nov 25 '15 at 03:37
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    @jameslarge: Your comment appears not to address the question at all. – WillO Nov 25 '15 at 05:26
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    It's not clear what is meant by "the most stationary on average". How are you computing the average? If one rock contains 1,000,000 times as many elementary particles than another, do you average over the two rocks, or over the kazillion elementary particles? – WillO Nov 25 '15 at 05:29
  • I am the most stationary object in the universe, and I can prove it experimentally. In fact, anyone can. – Asher Nov 25 '15 at 06:23
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    Not sure why there is such a bad reaction to this question. It's a natural thing to wonder about. Moreover, to everyone saying there's no way to compute something like this, it should be pointed out that there is a preferred reference frame in our universe, that of the CMB, and we have measured our motion with respect to it. –  Nov 25 '15 at 12:07
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    @ChrisWhite: I don't think OP is looking for a preferred reference frame, he's looking for a frame which is maximally at rest w.r.t. other frames. AFAIK, the CMB frame doesn't do this. – Kyle Kanos Nov 25 '15 at 13:17
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    Related: http://physics.stackexchange.com/q/25591/2451 – Qmechanic Nov 25 '15 at 14:32
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    @WillO, Comment deleted. Apparently I misread the question, but even so, does the reference frame that follows the center of mass of the cosmos have any special significance when compared to any other inertial reference frame? (I.e., Do the laws of physics take on a simpler, more fundamental form when expressed in that frame?) – Solomon Slow Nov 25 '15 at 14:39

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To make sense of this question you need to decide what an "object" means (is a rock an object or a conglomeration of a vast number of much smaller objects?).

Once you've settled that, you need to decide whether you're averaging velocities or speeds. If there are three objects, and two of them are moving away from me at the same speed $v$ in opposite directions, are those two objects, on average, stationary with respect to me or are they, on average, moving away from me at speed $v$? Since you want to talk about things like "the smallest", you presumably mean speed (which is a number) rather than velocity (which is not).

In a non-flat spacetime, there's also the issue of how you define the velocity relative to you of a very distant object with which you do not share a coordinate patch.

And finally, because velocities are always changing, we need to do this for all objects at a fixed time, which is going to require something like a global time coordinate.

But once you've come up with (necessarily pretty arbitrary) answers to all these questions and issues, the answer to your question is: It depends.

If there are finitely many objects, you can certainly do your averaging to associate a number to each one, and then choose the smallest of those numbers.

If there are infinitely many objects, it's less clear how to do the averaging. You'll need some kind of measure on the space of objects to integrate against. Once you figure out how to do that, you'll have associated a number to each object, and those numbers are bounded below (by zero) so they have a greatest lower bound. There might or might not be some object that realizes that bound, but there are certainly objects that get arbitrarily close.

Why you would ever want to do this is, of course, a much harder question.

Edited to add: Re your question about time dilation --- time dilation does not scale linearly with speed, so the object with the smallest average speed will probably not be the object with the smallest average time dilation factor. You could of course, in principle, identify that object --- but the reasons for doing so seem even more obscure than for the object with the smallest average speed.

WillO
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No, there cannot be a most stationary object in the universe. One of the basic tenets of relativity is that every inertial reference frame is equivalent. This is what Einstein said on the matter,

If a co-ordinate system $K$ be so chosen that when referred to it the physical laws hold in their simplest forms, these laws would be also valid when referred to another system of co-ordinates $K'$ which is subjected to an uniform translational motion relative to $K$.

(source)

If we have two objects, say a star & a comet, to continue with your astronomical objects, then we can create one coordinate system for the sun (call it $K$) and one for the comet (call it $K'$). In each of these coordinate systems, the respective object, sun or comet, is stationary (so the sun is stationary in $K$, the comet in $K'$). If $K$ and $K'$ differ (one sees the other moving), then the sun believes it is stationary and sees the comet as moving while the converse is true for the comet. And they're both right.

On the flip side, in order for these two frames to agree that the sun is stationary is for them to both be the same frame: $K'=K$. This is what physicists call co-moving (proper) frames.

You want to extrapolate this to find $n$ frames such that they are all described by coordinate system $C$. Given the $\sim10^{11}$ stars in the galaxy and (approximately) equal number of galaxies, all of which are moving in a variety of ways, so you will not be able to find such a 'most stationary' frame, it just doesn't exist.

Kyle Kanos
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    It seems to me that this answer is most certainly wrong, though because of ambiguity in the question it's not clear what answer is right. But to see that this is wrong, imagine a universe with 100 particles, 99 of them at rest with respect to each other and the last moving at velocity $v$ with respect to the 99. Then if we average any particle's 99 relative speeds, we're going to get $|v|$ for the odd particle out and $|v|/99$ for the others. So the answer in this case (on pretty much any interpretation) is: any of the $99$. – WillO Nov 25 '15 at 16:00
  • @WillO: Sure you can create an unrealistic counter-example that doesn't match observations/reality and call this answer wrong. – Kyle Kanos Nov 25 '15 at 16:06
  • But feel free to write your own answer that says such a frame exists, if you feel that it does. – Kyle Kanos Nov 25 '15 at 16:08
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    The question I raised was not whether such a frame exists but whether your answer rules it out. If your reasoning can't rule out this counterexample, then it can't prove non-existence. – WillO Nov 25 '15 at 17:31
  • The problem with your argument is that you are pre-supposing a near-uniform set of reference frames exists (with just 1 different from the others) and then later discovering that you have a near-uniform set of reference frames. Perhaps you should try not assuming the result first? – Kyle Kanos Nov 25 '15 at 17:58
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    But the near-uniform set of reference frames is not important to the counterexample. All I need is a bunch of objects, all moving respect to each other. To each object I associate the average relative speed of all the other objects. That gives me a bunch of numbers and I pick the smallest. I still don't see how your argument rules this out, though perhaps I'm missing something that seems to you to be too obvious to explain. – WillO Nov 25 '15 at 18:15
  • It seems to me that OP is looking for a reference frame, $K$ that the most amount of objects belong to. So your focusing on averaging (which I don't think OP meant in the sense you're taking it) is, to me, irrelevant. – Kyle Kanos Nov 25 '15 at 18:20
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    Ah! Now I understand why you think this is an answer. But I do believe you're mistaken about what the OP is asking for. Note that the original post asks for the object relative to which most objects are stationary on average. I think this means we're supposed to focus on averaging. – WillO Nov 25 '15 at 18:22
  • OP said on average so that you could ignore perturbations to the frame $K$ (e.g., ignoring earth's rotation when considering that it orbits the sun). Picking up on just a single word and ignoring the meaning is probably why you are so off-base here. – Kyle Kanos Nov 25 '15 at 18:34
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    @WillO : this answer is wrong. The reference frame you're referring to is the CMB rest frame, see this answer by ghoppe and note the quotes by Dough Scott. – John Duffield Dec 20 '15 at 15:44
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    @ohnDuffield: I think you might have misdirected your comment; it seems to have nothing to do with anything I was saying. Was it meant for someone else? – WillO Dec 20 '15 at 16:42