In Bestvina and Handel's paper on train track maps, they state that a pseudo-Anosov map on a punctured surface with one orbit of punctures induces an irreducible automorphism, which is not fully irreducible if there's more than one puncture (Example 1.4 on Page 6). They don't give a proof of this and I haven't been able to find a proof (nor a sketch of one) anywhere.
Suppose $p:\Sigma \to \Sigma$ was the homeomorphism, $\phi = [p_*]$ was the induced outer automorphism in $Out(F_n) \cong Out(\pi_1(\Sigma))$, and $f:\Gamma \to \Gamma$ was map of graphs that was a reduction for $\phi$, i.e., $\phi = [f_*]$ in $Out(F_n) \cong Out(\pi_1(\Gamma))$ and it has a non-contractible invariant proper subgraph. How am I supposed to relate $\Sigma$ and $\Gamma$ to get a contradiction? The two are homotopy equivalent but there's no reason for $\Gamma \to \Sigma$ to be an embedding.
In that paper, the definition of an irreducible automorphism is one that does not fix (the conjugacy class of) any proper free factor. This is equivalent to the invariant subgraph definition, but for this example it's easier to think about free factors. Suppose $\Sigma$ has $k$ punctures and choose a basis for $\pi_1(\Sigma)$ that contains a peripheral curve (a curve $c$ such that one component of $\Sigma-c$ is a once-punctured disk). Then $\phi(c)\neq c$ since $f$ induces a non-trivial permutation of the punctures, but $\phi^{k!}(c)=c$ as the permutation of the punctures has order dividing $k!$. The point is that $[c]$ is a free factor, since $c$ is primitive. Since $f$ is pseudo-Anosov, the only (homotopy classes of closed) curves on the surface that are fixed by $\phi$ are peripheral ones.