Let $N \in \textrm{Mat}_n(\mathbb C)$ be a nilpotent matrix. Let $g \in \operatorname{GL}_n(\mathbb C)$ be an invertible matrix, and let $g = g_sg_u$ be the multiplicative Jordan decomposition of $g$. That is, $g_s$ is diagonalizable, $g_u$ is unipotent, and $g_sg_u = g_ug_s$. Suppose $N$ is an eigenvalue for the linear operator $\operatorname{Ad}g$ on $\textrm{Mat}_n(\mathbb C)$, which is to say $gNg^{-1} = \lambda N$ for some $0 \neq \lambda$.
Does $g_u$ then necessarily commute with $N$?
I think this comes down to the following: let $T: \textrm{GL}(V) \rightarrow \textrm{GL}(V)$ be an invertible linear transformation of a finite dimensional complex vector space. Let $T_sT_u$ be the multiplicative Jordan decomposition of $T$. If $v \in V$ is an eigenvector for $T$ with eigenvalue $\lambda$, then it is also an eigenvector for $T_s$ with the same eigenvalue $\lambda$. Hence $T_uv = v$.
This comes down to looking at the generalized eigenspace $$V_{\lambda} = \{w \in V : (T - \lambda 1_V)^kv = 0 \textrm{ for some } k \geq 0 \}$$