  Let $N \geq 2$ and let $\mathrm{Out}(F_N)$ be the outer automorphism group of
a nonabelian free group of rank $N$. Let
$\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ be the finite index subgroup of
$\mathrm{Out}(F_N)$ which is the kernel of the natural action of
$\mathrm{Out}(F_N)$ on $H_1(F_N,\mathbb{Z}/3\mathbb{Z})$. We show that
$\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ is an $R$-group, that is, for every
$\phi,\psi \in \mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$, if there exists $k \in
\mathbb{N}^*$ such that $\phi^k=\psi^k$, then $\phi=\psi$. This answers a
question of Handel and Mosher. We then use the fact that
$\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ is an $R$-group in order to prove that
the normalizer in $\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ of every abelian
subgroup of $\mathrm{IA}_N(\mathbb{Z}/3\mathbb{Z})$ is equal to its
centralizer. We finally give an alternative proof of a result, due to Feighn
and Handel, that the centralizer of an element of $\mathrm{Out}(F_N)$ which has
only finitely many periodic orbits of conjugacy classes of maximal cyclic
subgroups of $F_N$ is virtually abelian.
