Denote by $F\_n$ the free group of rank $n$. We say that an automorphism $\phi\in Aut(F\_n)$ is *geometric* if there exists a surface with boundary $M$ and a homeomorphism $h\colon M\to M$ such that $h$ induces $\phi$ on $\pi\_1$.


Every automorphism of $F\_2$ is geometric, but in higher rank geometric automorphisms are rare. I have seen several examples of nongeometric automorphisms, but all of which were of infinite order.


**Question:** Are there any examples of *periodic* (i.e. finite-order) nongeometric automorphism? Or can one show that periodic automorphisms are always geometric, possibly making use of the fact that every periodic automorphism can be realized as a simplicial graph automorphism? I wouldn't be surprised if we could nicely embed the graph into a suitable surface and extend the automorphism of the graph to a surface homeomorphism.


**Observation:** Every periodic automorphism $\phi$ fixes some nontrivial conjugacy class (maybe up to inversion), which is a necessary condition for $\phi$ to possibly be induced by a homeomorphism of a surface with *connected* boundary. But this observation really only plays a role when $n$ is even, as the fundamental group of a surface with connected boundary has even rank.



Periodic automorphisms of free groups and surface homeomorphisms
Assuming that you mean outer automorphisms of $F\_n$ (as per my comment), the answer is that if $n \ge 3$ then there is a nongeometric finite order element of $Out(F\_n)$, for example the outer automorphism class of $a\_1 \mapsto a\_1^{-1}$ and $a\_i \mapsto a\_i$ for $2 \le i \le n$. 


For the proof, given $h : M \to M$ that induces a nonidentity, finite order, outer automorphism $\phi$ of $\pi\_1 M$, I'll show that there does not exist a free factor of $\pi\_1 M$ of rank $\ge 2$ whose conjugacy class is fixed by $\phi$ and whose elements all have conjugacy classes fixed by $\phi$, in contrast with the above example having such a free factor of rank $n-1$.


By Nielsen realization, after an isotopy of $h$ we may assume that $h$ is a finite order isometry with respect to some hyperbolic structure on $M$ with totally geodesic boundary. If there is a free factor $G < \pi\_1 M$ of rank $\ge 2$ with conjugacy class fixed by $h$ then $h$ lifts to the $G$-cover $\tilde M$, and that lift restricts to an isometry of the convex hull $\tilde h : \mathcal{H}(\tilde M) \to \mathcal{H}(\tilde M)$. Since $rank(G) \ge 2$, the convex hull $\mathcal{H}(\tilde M)$ is itself a compact hyperbolic surface with totally geodesic boundary (if $rank(G)$ were equal to $1$ then the convex hull would just be a closed geodesic, and no contradiction would arise). Since $h$ fixes the conjugacy classes of all the elements of $G$, and since $G$ is a free factor (malnormality of $G$ suffices), it follows that $h$ induces the identity outer automorphism of $\pi\_1(G)$. From this it follows in turn that $\tilde h$ is the identity isometry on $\mathcal{H}(\tilde M)$. It follows by projection that $h$ is locally the identity at some points of $M$, which by analytic continuation implies that $h$ is globally the identity. So $h$ induces the identity outer automorphism.


