TITLE: growth of a free group automorphism is same for finite index subgroups?
QUESTION [7 upvotes]: 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 $\phi$.)

REPLY [3 votes]: An alternate way to address point 5 in Lee Mosher's above answer, using only that $\phi$ and $\phi|_H$ have quasi-isometric mapping tori, is to use a result of Macura (Macura, Nataša, Detour functions and quasi-isometries., Q. J. Math. 53, No. 2, 207-239 (2002). ZBL1036.20033.).
Indeed, let $G=F\rtimes_\phi\mathbb Z$ and let $G'=H\rtimes_\phi\mathbb Z$.  Suppose that $\phi$ has polynomial growth of order $r$.  Then, as Lee explained above, $\phi|_H$ also has non-exponential, and thus polynomial, growth.  But Macura's result (Theorem 1.1 in the above) says that $G$ has detour function a polynomial of degree exactly $r+1$, and another application of the same theorem shows that $G'$ has detour function a polynomial of degree $r'+1$, where $r'$ is the degree of the polynomial growth function of $\phi|_H$.  But $G$ is obviously quasi-isometric to $G'$, and Macura shows that the order of the detour function is a quasi-isometry invariant, so $r=r'$.    
(The detour function is very similar to the perhaps more familiar divergence function.)