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Crossposted on Mathematics.

In this post, an order-preserving self-map of a poset $X$ will be called an endomorphism of $X$, and such an endomorphism $f$ will be called rigid if the only automorphism of $X$ which commutes with $f$ is the identity of $X$.

The question is in the title:

Does every finite poset have a rigid endomorphism?

Every poset of cardinality at most $9$ has a rigid endomorphism.

I wrote a proof of this statement in a separate text. Since links tend to break over time I am including several links to this text:

pdf file --- tex file --- Overleaf --- Google Drive --- Mediafire.

In the first version of this question I put the proof in the post itself. But I realized that there was a mistake, and that the post was too long. So I rewrote the proof (hoping that it is correct now), and added links to the new version.

Pierre-Yves Gaillard
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    related (while distinct) to https://mathoverflow.net/questions/358057/for-what-sets-x-do-there-exist-a-pair-of-functions-from-x-to-x-with-the-id, https://mathoverflow.net/questions/359660/centralizer-of-a-single-element-in-the-monoid-of-self-maps-of-a-set – YCor Oct 11 '20 at 13:44
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    This looks like a very interesting question (although the wall of text is a bit intimidating- I wish MO had a way of putting stuff inside spoiler tags). But one naive question: do you have a counter-example for infinite posets (where I think everything still makes perfect sense)? – Sam Hopkins Oct 11 '20 at 13:47
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    Note that the question is whether the monoid $M$ of endomorphisms of every finite poset satisfies the purely monoid-wise property: $$\exists g\in M:\forall h,k\in M: (hk=kh=1 \text{ and } hg=gh) \Rightarrow h=1.$$ – YCor Oct 11 '20 at 13:49
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    @SamHopkins- Thanks! No I don't have a counterexample for infinite posets, but I'm very interested in the infinite case as well. I thought it was natural to concentrate first on the finite case. – Pierre-Yves Gaillard Oct 11 '20 at 14:02
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    Any chain is the image of an endomorphism, so if you can find a chain which isn't fixed by any nontrivial automorphism you're done. (I have no intuition for whether this should be expected to exist in general, but many highly symmetric posets I can think of have this property, eg. any maximal chain in a boolean lattice.) – lambda Oct 21 '20 at 14:09
  • To pedantically correct myself, I meant any nonempty chain. – lambda Oct 21 '20 at 14:24
  • @lambda - Thanks a lot! Looks very interesting! If I understand correctly you're saying that if $C$ is a maximal chain in a boolean lattice, then $C$ isn't fixed by any nontrivial automorphism. Could you give me an argument or a reference? – Pierre-Yves Gaillard Oct 21 '20 at 15:49
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    @Pierre-YvesGaillard The boolean lattice $B_n$ consists of subsets of ${1, \dots, n}$ and its automorphism group is $S_n$ acting in the obvious way. A maximal chain looks like ${X_0, X_1, \dots, X_n}$ where $X_{i+1}$ is obtained by adding a single element to $X_i$. Any element of $S_n$ that fixes both $X_i$ and $X_{i+1}$ must therefore fix this new element, so an automorphism that fixes the whole chain is the identity. – lambda Oct 21 '20 at 16:17

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