Permuting subgroups with the same finite index Suppose that we have a finitely generated residually finite group $G = \langle g_1,\ldots,g_r \rangle$ with polynomial growth. Let $H$ be a subgroup of $G$ with finite index $m$. Let $\phi$ be an automorphism on $G$.
Question: What is the bound for the smallest $n \in \mathbb{N} \setminus \{0\}$, such that $\phi ^n(H) = H$?
My thought so far: We know that $G$ has at most $(m!)^r$ number of subgroups with index $m$, so $n \leq (m!)^r$. Are there any known results we can use to improve this bound? Is it possible to show that $n$ is bounded by a polynomial (or even linear function) in $m$, i.e. $ n = \mathcal{O}_r(m^d)$ for some $d$?
If $G$ doesn't have polynomial growth, then $n$ might not be bounded by a polynomial function in $m$ because of this example.
 A: The main result of
Lubotzky, Alexander; Mann, Avinoam; Segal, Dan, Finitely generated groups of polynomial subgroup growth, Isr. J. Math. 82, No. 1-3, 363-371 (1993). ZBL0811.20027.
states that a finitely generated, residually finite group has polynomial subgroup growth if and only if it is virtually solvable.
Therefore, virtually nilpotent groups have polynomial subgroup growth, that is, your question if a polynomial bound exists has a positive answer.
(Comment. Your question was posed for finitely generated groups of polynomial growth. By a well known result of Gromov, this is precisely the class of finitely generated, virtually nilpotent groups. Also, finitely generated nilpotent groups are always residually finite, so the assumption that the group is residually finite is not needed, it is automatically satisfied).
As for linear bounds, this is too restrictive and the class of groups for which the subgroup growth function $s_n(G)$ satisfies
$$ \limsup \frac{\log s_n(G)}{\log n} = 1$$
is fairly limited and fully described in
Shalev, Aner, On the degree of groups of polynomial subgroup growth, Trans. Am. Math. Soc. 351, No. 9, 3793-3822 (1999). ZBL0936.20021.
