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My question is a little difficult to explain, but I will try to state it. First, let me talk about the field $\Bbb R$ of real numbers. Let $K$ be the maximal real algebraic extension of $\Bbb Q$, that is, the maximal subfield of $\Bbb R$ which is algebraic over $\Bbb Q$. We have that the group of all automorphisms of $K$ is very small (!) in the sense that we do not have a lot of options to choose it: for example, any automorphism $f$ of $K$ must satisfy $f(\sqrt{d})=\sqrt{d}$ for all square-free positive integer $d$ (because every square is non-negative in $\Bbb R$). Under this context, I may say that $\Bbb R$ is algebraically restrictive.

My question is the following: Given a prime number $p$, does the $p$-adic field $\Bbb Q_p$ have an algebraic restriction in the sense that we can't choose freely an automorphism of the maximal subfield of $\Bbb Q_p$ which is algebraic over $\Bbb Q$? If it happens, how "big" is this restriction of $\Bbb Q_p$?

Ángel Valencia
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    Using squares we can show that neither $\Bbb{R}$ nor $\Bbb{Q}_p$ themselves have any non-trivial automorphisms. Not sure to what extent this restricts the automorphisms of maximal algebraic subfields. See this thread for the basic arguments. May be the links given there lead to something that helps you? – Jyrki Lahtonen Feb 02 '16 at 21:35
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    @JyrkiLahtonen, thank you so much for your hint. Using the arguments of the thread, you have that there exists only one automorphism of $\overline{\Bbb Q}\cap\Bbb Q_p$ for all place $p$ (I stand $\Bbb Q_\infty=\Bbb R$). – Ángel Valencia Feb 25 '16 at 09:39

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