There is a Nielsen-Thurston type method due to Feighn and Handel which is useful for approaching this question. The method is laid out in the papers arXiv:math/0612702 and arXiv:math.GR/0612705 by Feighn and Handel, "The recognition theorem for $Out(F_n)$ and "Abelian subgroups of $Out(F_n)$". It is an outgrowth of the train track and relative train track machinery developed in earlier papers of Bestvina, Feighn, and Handel, particularly math.GT/9712217, part I of their series on the Tits alternative for $Out(F_n)$.

As Ashot suggests, take $F_4 = \langle a,b,c,d \rangle$ and take $\Phi \in Aut(F_4)$ to preserve $\langle a,b,c\rangle $ so that the action on $\langle a,b,c\rangle $ has no periodic conjugacy classes. This implies that $\Phi(d) = u d^{\pm 1} v$ for words $u,v$ in $a,b,c$. One could take $\Phi(d)=d$ as Ashot suggests but I prefer $u$ and/or $v$ to be nontrivial for reasons explained below. Any such $\Phi$ is a "principal automorphism", which is guaranteed by having a fixed subgroup (when $u$, $v$ are trivial) or by having no fixed subgroup but having three or more attracting periodic points in the boundary of the free group (when $u$ and/or $v$ are nontrivial). Furthermore, if you throw in the condition that the action of $\Phi$ on the $a,b,c$ rose is a train track map and has no periodic Nielsen paths, then $\Phi$ is the ONLY principal automorphism in its outer automorphism class $\phi \in Out(F_n)$ up to conjugation by inner automorphism.

The centralizer $C(\phi)$ in $Out(F_n)$ has to act on the "principal data" of $\phi$, meaning that it has to permute the conjugacy classes of fixed subgroups, and it has to permute orbits of attracting periodic points in the boundary. Furthermore, $\phi$ has an attracting lamination $\Lambda$ that is supported in the $a,b,c$ subgroup, and there is an association between certain attracting periodic points in the boundary and certain rays of the lamination $\Lambda$, and this association implies that $C(\phi)$ preserves $\Lambda$ and so has a well-defined stretch factor homomorphism for $\Lambda$. The gist of the Recognition Theorem in this situation is that an element of $C(\phi)$ is more-or-less determined by all of this "principal data". The situation is a little more complicated to analyze when there is a nontrivial fixed subgroup, so for that reason I actually prefer the opposite case where $u$ and/or $v$ is nontrivial, which for the class of examples under consideration implies that there is no fixed subgroup for any automorphism representing $\phi$ (this is one place where $Out(F_n)$ departs from $MCG(S)$, for if a mapping class is not pseudo-Anosov then after passing to a power there MUST be a principal automorphism having nontrivial fixed subgroup). So in that case, $C(\phi)$ acts on the set of orbits of attracting periodic points, this is a finite set, and so the action has a finite index kernel; the kernel of that action has a stretch factor homomorphism to $Z$, and the kernel of THAT homomorphism is trivial.

`$Out(F_n)$`

is irreducible if it does not fix (the conjugacy class of) some free factor`$F<F_n$`

(i.e.`$F_n=F*G$`

for some other subgroup`$G$`

). It is irreducible with irreducible powers (iwip) if its powers have the same property as well. – Alessandro Sisto Dec 7 '11 at 21:34