Hi, 
I am troubled with the following question: Does there exist a finite order automorphism of a free group, $f\in Aut(F)$, such that it fixes no non trivial conjugacy class and no non trivial centralizer, i.e. $f(g)$ is not conjugate to $g$ and $f(g)\neq g^{-1}$ for any $g\in F$ ? Can we find such, so the same holds for any of its non trivial powers $f^i$?


By inspection of the finite order automorphisms of $F\_2$, this cannot be an automorphism of $F\_2$. 


Moreover, if there exists such an automorphism is it possible to find one with as large order as we want (possibly after moving to a larger free group)? 



Periodic automorphisms of free groups
No such automorphism exists. Every finite order automorphism of a finite rank free group has a nontrivial fixed conjugacy class. To see why, you can represent the free group automorphism as a simplicial automorphism $f : G \to G$ of some finite connected graph having no vertices of valence 1. Take any vertex $p \in G$ of valence $\ge 3$. Let $\gamma$ be an immersed edge path with initial oriented edge $E\_0$ having initial vertex $p$, and terminal oriented edge $E\_1$ having terminal vertex $f(p)$, such that the initial direction of $f(E\_0)$ does not equal the terminal direction of $E\_1$. The valence condition on $p$ lets you make this choice; if you made a bad choice of $\gamma$ then, since the valence of $f(p)$ is $\ge 3$, you could concatenate with some immersed path from $f(p)$ to $f(p)$ so as to change your terminal direction. Letting $k$ be the order of the simplicial automorphism $f$, it follows that $f$ fixes the conjugacy class corresponding to the immersed loop $\gamma \* f(\gamma) \* ... \* f^{k-1}(\gamma)$. 


