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 $\phi$.)



growth of a free group automorphism is same for finite index subgroups?
While I cannot point to any place in the literature with this particular statement, I would say that it is an exercise in relative train track theory, which is the very nice normal form theory for $\text{Out}(F\_n)$ and $\text{Aut}(F\_n)$ found in papers of Bestvina, Feighn, and Handel (in various subgroupings). Usually relative train track theory is applied to $\text{Out}(F\_n)$ but you can apply it to $\text{Aut}(F\_n)$ with a little trick.


First, you can embed $\text{Aut}(F\_n) \hookrightarrow \text{Out}(F\_{n+1})$ by letting $F\_{n+1}=\langle a\_0,a\_1,...,a\_n\rangle$, and restricting to the subgroup in which $a\_0$ is fixed and $a\_1,...,a\_n$ are mapped to themselves. Every $\phi \in \text{Aut}(F\_n)$ is represented by a relative train track map $f : G \to G$ where $G$ has a closed edge labelled $a\_0$ with base point $p$ such that $a\_0$ is fixed by $f$, and $f$ preserves $G \setminus H$, inducing the automorphism $\phi$ on $\pi\_1(G \setminus H,p) \approx F\_n=\langle a\_1,...,a\_n\rangle$. Let's refer to this map $f$ as a "relative train track representative of $\phi \in \text{Aut}(F\_n)$" (this is a nonstandard terminology: ordinarily relative train maps are things which represent outer automorphisms, and this $\phi$ does indeed represent an outer automorphism of $F\_{n+1}$, but we have used the "fixed loop trick" to make $\phi$ also represent an actual automorphism of $F\_n$).


I still have to say what the "normal form" properties of $f$ are, i.e. I have to tell you a little bit about the definition of a relative train track map. After passing to a power, $G$ is written as a nested union
$$G\_0 \subset G\_1 \subset ... \subset G\_K=G
$$
of $f$-invariant subgraphs, starting with $G\_0=a\_0$. For each subgraph $H\_i = G\_i - G\_{i-1}$ there are three possibilities: $H\_i$ is a "zero stratum" which means that $H\_i$ is contractible and its image under $f$ is contained in $G\_{i-1}$; or $H\_i$ is an "NEG stratum" (meaning "non-exponentially growing") meaning that $H\_i$ is a single edge $E$ such that $f(E)=Eu$ with $u$ some path in $G\_{i-1}$; or $H\_i$ is an "EG stratum" meaning that has $m \ge 2$ edges the $m \times m$ transition matrix of $f$ acting on the edges of $H\_i$ is a Perron-Frobenius matrix, and some additional "non cancellation" properties hold which I won't bother to make explicit. 


So, the major points are:


1. One can detect the growth type of $f$ from this normal form. For example, $\phi$ has exponential growth if and only if some stratum $H\_i$ is an EG stratum. The NEG strata are so named because each crosses itself only once, but the NEG strata can generate polynomial growth by the manner in which they cross over lower stratum edges (think of an example on $F\_3 = \langle a\_1,a\_2,a\_3 \rangle$ where $a\_1 \mapsto a\_1$, $a\_2 \mapsto a\_2 a\_1$, and $a\_3 \mapsto a\_3 a\_2$; in this example, $a\_3$ has quadratic growth). One outcome of this is that every automorphism of $F\_n$ has either exponential growth or polynomial growth of some integer degree (between $0$ and $n-1$).
2. One can obtain a relative train track map representing $\phi | H \in \text{Aut}(H)$ from a covering space trick as follows. Construct a pointed covering space of $(G \setminus a\_0,p)$ representing the subgroup $H$. Then attach a closed edge to each lift of $p\_0$, obtaining a covering space $G'$ of $G$, a subgraph $A \subset G'$ which is the union of the lifts of $a\_0$, and a base point $p' \in G'$ to which one of the loops $a'\_0$ is attached, so that $(G' \setminus A,p')$ is the covering space of $(G \setminus a\_0,p)$ representing $H$. The map $f$ lifts to a map $f' : G' \to G'$ fixing $p'$, and preserving $G' \setminus A$, so that the restiction of $f'$ to $(G' \setminus A,p')$ induces the homomorphism $\phi | H$. Thus, we may think of $f' : G' \to G'$ as a train track representative of $\phi | H$ (again a nonstandard terminology). One needs to check that $f'$ satisfies the relative train track axioms, using that $f$ satisfies those axioms, but this is straightforward. We can therefore use $f'$ to detect the growth properties of $\phi | H$.
3. If $\phi$ has exponential growth, then $G$ has an EG stratum $H\_i=G\_i \setminus G\_{i-1}$, implying that the inverse image of $H\_i$ in $G'$ is a (union of) EG strata of $f'$, implying that $\phi | H$ has exponential growth.
4. If $\phi$ has polynomial growth, equivalently $\phi$ does not have exponential growth, then every stratum of $G$ is an NEG stratum, implying that every stratum of $G'$ is an NEG stratum, implying that $\phi | H$ does not have exponential growth, implying that $\phi | H$ has polynomial growth.
5. I believe that one can argue similarly regarding degrees of polynomial growth, demonstrating that if $\phi$ has polynomial growth of degree $d$ then $\phi'$ has polynomial growth of degree $d$.


