
> 
> Is there a profinite group $G$, a continuous automorphism $\alpha$ of
>  $G$ and a topologically finitely generated closed subgroup $H \leq G$
>  such that $\alpha(H) \lneq H$ ?
> 
> 
> 


Note that if an example exists, then $G$ is not topologically finite generated, and $\alpha$ is not given by conjugation by an element of $G$.


If we allow $H$ not to be topologically finitely generated, then the example given by Yves de-Cornullier in [Wild automorphisms of a free group](https://mathoverflow.net/questions/224360/wild-automorphisms-of-a-free-group) gives rise to an example here.



Wild automorphisms of profinite groups
No.


Let $K\_n = K\_n(G)$ be the intersection of all open normal subgroups of $G$ of index at most $n$. Then $\alpha(K\_n) = K\_n$. If we replace $G$ with $G/K\_n$, we still have $\alpha$ acting as an automorphism on $G/K\_n$ and for $n$ large enough, $\alpha(H)K\_n/K\_n$ is a proper subgroup of $HK\_n/K\_n$ (since $H$ and $\alpha(H)$ are both closed, we can distinguish them using finite images of $G$). So we may assume $K\_n(G)$ is trivial for some $n$. It then follows that $K\_n(H)$ is trivial. But if a topologically finitely generated profinite group $H$ satisfies $K\_n(H) = 1$ for some $n$, then $H$ must be finite, since $H$ only has finitely many finite quotients of any given order. This gives a contradiction.


