Eigenvectors of an isomorphism. Let $f: \mathbb{R}^{n} \to \mathbb{R}^{n}$ be a linear isomorphism, with $n \geq 4$. I have to prove that $f$ and $f^{3}$ have the same eigenvectors.
Clearly, every eigenvector of $f$ is an eigenvector of $f^{3}$. For the converse, I think that if $v$ is an eigenvector of $f^{3}$ associated to the eigenvalue $\lambda \neq 0$ then $v$ is also a eigenvector of $f$ associated to the eigenvalue $\sqrt[3]{\lambda}$. But I am lost trying to prove this.
Thanks in advance.

Let $n=4$ and $f(v)=Av$, where $A$ is a rotation by 120° degrees in the first two coordinates, i.e.
$$A=\begin{pmatrix}\cos(2\pi/3) & \sin(2\pi/3) & 0 & 0\\-\sin(2\pi/3) & \cos(2\pi/3)&0&0\\ 0&0&1&0\\ 0&0&0&1\end{pmatrix}.$$
I'm pretty sure that $f$ has two eigenvectors, which are the third and fourth unit vector, while $f^3$ is the identity and thus has four eigenvectors, namely the unit vectors.
