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Let $G$ be a group with a subgroup $H$ such that the index of $H$ is $n$. For any automorphism $\tau$ of $G$, can we prove that $\tau^{n!}$ preserves $H$? Here $\tau^{n!}$ means the self-composition of $\tau$ for $n!$ times.

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$\tau$ acts on the set of subgroups of index $n$. There can be many such subgroups, so we cannot expect something like you suggest. Indeed, let $$ G = (\Bbb Z/n\Bbb Z)^M$$ and let $H$ be the kernel of projection to the first factor, and let $\tau$ be cyclic permutation of the $M$ components. Then the order of $\tau$ in $\operatorname{Aut}(G)$ is $M$ (and for no smaller power is $H$ preserved), which may be $>n!$.

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The smallest counterexample is the following: let $G = \mathbb{Z}_2 \oplus\mathbb{Z}_2$, $H = \mathbb{Z}_2 \oplus 0 = \langle (1,0)\rangle$ and $$\tau: \mathbb{Z}_2 \oplus \mathbb{Z}_2 \to \mathbb{Z}_2 \oplus \mathbb{Z}_2: (x,y) \mapsto (y,x+y).$$ Then $\tau(H) = 0 \oplus \mathbb{Z}_2 = \langle (0,1)\rangle$ and $\tau^2(H) = \langle (1,1)\rangle \neq H.$

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