# Automorphisms of Free Groups of Finite Order

The automorphism group of a free group (of finite rank) is known (see this). The group is infinite, if the number of generators is at least $2$, since there are automorphisms of the form $x_i\mapsto x_ix_j$ and other $x_k$'s fixed (here $\{x_i\}_{i=1}^n$ is a minimal generating set). Also note that $S_n$ is a subgroup of automorphism group of free group of rank $n$.

Question: Consider the automorphisms of a free group of rank $n$ which are of finite order; is there upper bound on this order in terms of $n$?

Yes, there is such a bound, namely $$n! 2^n$$. You can look this up in Vogtman's survey paper entitled "Automorphisms of free groups and outer space". Here's an outline.
Let's denote the rank $$n$$ free group by $$F_n$$ and its automorphism group by $$\text{Aut}(F_n)$$. Also, denote its normal subgroup of inner automorphisms by $$\text{Inn}(F_n)$$, and its outer automorphism group $$\text{Out}(F_n) = \text{Aut}(F_n) / \text{Inn}(F_n)$$. Since $$\text{Inn}(F_n)$$ is isomorphic to $$F_n$$ itself which is torsion-free, for any finite subgroup $$G \subset \text{Aut}(F_n)$$ the quotient homomorphism $$\text{Aut}(F_n) \mapsto \text{Out}(F_n)$$ is injective on $$G$$. It therefore suffices to bound the orders of finite subgroups of $$\text{Out}(F_n)$$.
The theorem you need is a realization theorem (proved independently by Culler, by Khramtsov, and by Zimmerman). This theorem says that for every finite subgroup $$G < \text{Out}(F_n)$$ there exists a connected, finite graph $$\Gamma$$ whose vertices all have valence $$\ge 3$$ and whose fundamental group is free of rank $$n$$, such that $$G$$ is isomorphic to a subgroup of the simplicial automorphism group of $$\Gamma$$. Since the number of vertices and edges of $$\Gamma$$ is bounded in terms of $$n$$ (by a simple Euler characteristic calculation), one obtains a concrete bound on the order of the simplicial automorphism group of $$\Gamma$$ and therefore on the order of $$G$$.
The largest finite order subgroup of $$\text{Out}(F_n)$$ is the simplicial automorphism group of the "rose with $$n$$ petals", the graph with one vertex and $$n$$ edges each having both ends attached to the vertex. The automorphism group of this graph has order $$n! 2^n$$. It can be understood quite concretely in terms of a free basis $$F_n = \langle a_1,...,a_n \rangle$$, namely the group of automorphisms of the form $$a_i \mapsto a^{\epsilon(i)}_{\sigma(i)}$$ where $$\epsilon : \{1,...,n\} \to \{-1,+1\}$$ is any function and $$\sigma : \{1,...,n\} \to \{1,...,n\}$$ is any permutation. Up to isomorphism this is just the wreath product of $$\mathbb{Z}/2\mathbb{Z}$$ and the symmetric group $$S_n$$.