Let $V$ be a finite dimensional vector space over $\mathbb{C}$. Consider an automorphism $\Phi \in \text{Aut}(V)$,an endomorphism $N \in \text{End}(V)$ such that
$$\Phi N-\lambda N \Phi=0_{End_V}$$
for a complex number $\lambda \in \mathbb{C}$ with $|\lambda|<1$. I have shown that under these hypotheses, $N$ is nilpotent. My proof goes like this:
If $\lambda =0$, the claim is trivial since $\Phi$ is bijective: obviously $N$ must be the zero matrix. We know that $N$ is nilpotent if and only if its only eigenvalue is $0$. Let $\alpha$ be an eigenvalue w.r.t. the eigenvector $v$. Then
$$\lambda N \Phi (v)= \Phi N(v)= \Phi(\alpha v)= \alpha \Phi (v)$$
so $\frac{\alpha}{\lambda}$ is an eigenvalue of $N$ w.r.t. the eigenvector $\Phi (v)$. Through induction on the integer $n$ we can show that $\Phi^n (v)$ is an eigenvector w.r.t. the eigenvalue $\frac{\alpha}{\lambda ^n}$. If $\alpha \neq 0$, there would exist infinite eigenvectors w.r.t. to different eigenvalues. This cannot happen in a finite-dimensional space such as $V$, because eigenvectors relative to different eigenvalues are linearly independent, so $\alpha = 0$
Is there a more direct and easier way to prove this fact? I feel like demonstrating it this way is overcomplicated.
Since $\Phi$ is invertible we obtain $\Phi N\Phi^{-1}=\lambda N$ and hence $$ (\Phi N\Phi^{-1})^k=\Phi N^k\Phi^{-1}=\lambda^kN^k $$ for all $k\ge 0$. Hence we obtain $tr(N^k)=tr(\Phi N^k\Phi^{-1})=\dim(V)\lambda^{k}tr(N^k)$, which implies $tr(N^k)=0$ for all $k\ge x_0$. It is well-known that this implies that $N$ is nilpotent (over a field of characteristic zero).