Let $X=\{x_1,\dots,x_N\}$ and $F=F(X)$ be a free group generated by $X$. Let $\phi\colon F\to F$ be an automorphism of $F$. Define a growth function of $\phi$ as: $$ \operatorname{gr}_{\phi,X}(n)=\operatorname{max}_{1\le i\le N}\{\|\phi^n(x_i)\|_X\}, $$ where $\|.\|_X$ denotes the word length with respect to $X$. We consider these functions up to equivalence $f\simeq g$ defined in the following way. For functions $f,g\colon [0,+\infty)\to[0,+\infty)$ we say that $f \preceq g$ if there exist $C>0$ such that for all $n\in [0,+\infty)$: $$ f(n)\le Cg(Cn+C)+C. $$ We say that $f\simeq g$ if $f\preceq g$ and $g\preceq f$. We extend this relation to functions $\mathbb N\to [0,+\infty)$ by assuming them to be constant on each interval $[n,n+1)$.

[EDIT 07/28/17: Since the function $n\mapsto \|\phi^n(x_i)\|_X$ is not monotone (think of a finite order automorphism modeled on some subset of $X$), it is better to use a slightly different definition of $f\preceq g$: we say that $f\preceq g$ if there exist constants $A,B>0$, $C,D\geq0$ such that for all $n\in [0,+\infty)$: $$ f(n)\le Ag(Bn+C)+D.] $$

It can be shown that, viewed up to $\simeq$ equivalence, functions $\operatorname{gr}_{\phi,X}(n)$ are independent of generating set $X$ (so can be denoted just $\operatorname{gr}_\phi(n)$). And it looks plausible that:

**If $H\le F$ is a subgroup of finite index, invariant under $\phi$, then $\operatorname{gr}_{\phi|_H}(n)\simeq \operatorname{gr}_{\phi}(n)$.**

**Question:** Is the detailed proof of this statement written somewhere in the literature?

(I somehow find it difficult to prove that $\operatorname{gr}_{\phi|_H}(n)\succeq \operatorname{gr}_{\phi}(n)$, for arbitrary automorphism $\phi$.)