Let $F\_n$ denote the free group on $n$ generators $x\_1,\ldots , x\_n$.


Recall that an element $\varphi\in\mathrm{Aut}(F\_n)$ is an *IA automorphism* if it induces the identity on the abelianization $F\_n/[F\_n,F\_n]=\mathbb{Z}^n$. In other words, $\varphi$ is in the kernel of the induced map $\mathrm{Aut}(F\_n)\to Gl(n,\mathbb{Z})$. This kernel is denoted $\mathrm{IA}\_n$.



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> Does every $\varphi\in\mathrm{IA}\_n$ have a fixed point in the complement of the commutator subgroup $[F\_n,F\_n]$? 
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Fixed points of IA automorphisms
Here's a compilation of some of the answers given in comments.


As pointed out by Mark Sapir and Misha, if $S$ is a compact surface with one boundary component whose fundamental group is equipped with an isomorphism $\alpha : \pi\_1(S) \to F\_n$, and if $f : S \to S$ is any pseudo-Anosov homeomorphism, then the outer automorphism of $\pi\_1(S)$ induced by $f$ ($\leftrightarrow$ the mapping class of $f$) pushes forward via $\alpha$ to give an outer automorphism $\phi$ of $F\_n$ whose only periodic conjugacy classes are those associated to the boundary component of $S$. So for any automorphism representing $\phi$, all of its fixed points are in the commutator subgroup. And one can easily construct such a pseudo-Anosov $f$ which is in the Torelli subgroup of the mapping class group of $F\_n$: apply Penner's recipe, taking two filling, separating curves $c,d$ and composing a positive Dehn twist about $c$ with a negative Dehn twist about $d$. The (outer) automorphism of $F\_n$ induced by such an $f$ is in $IA\_n$.


But from those examples, one can construct further $IA\_n$ outer automorphisms whose only periodic conjugacy class is the identity. To do this, pick two isomorphisms $\alpha\_1,\alpha\_2 : \pi\_1 S \to F\_n$ taking the conjugacy class represented by $\partial S$ to two distinct conjugacy classes in $F\_n$. Let $\phi\_1,\phi\_2$ be the two $IA\_n$ outer automorphisms of $F\_n$ obtained by pushing $f\_\*$ forward via $\alpha\_1,\alpha\_2$ respectively, so their fixed conjugacy classes are distinct from each other. Then for high powers of the exponents, the $IA\_n$ outer automorphism $\phi\_1^i \phi\_2^j$ has no periodic conjugacy class at all; this is a consequence of Theorem 1 of [this recent preprint of Pritam Ghosh](https://arxiv.org/abs/1306.6049).


