I need to prove that if $\rho(A) = \rho(B)$ and $\rho(A^2) = \rho(B^2)$ then $A$ is similar to $B$. Assume also that $A,B$ are of dimension 9 and $\rho$ is the spectral radius and that $A,B$ are nilpotent with the same nilpotence index.
I tried to show they are diagonalizable and have the same diagonal but having problems finding the general form without knowing all the eigenvalues. Can we prove it this way? Or by any other way?
Even with the additional conditions, this doesn't seem to be true.
Consider $$ A=\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ and $$ B=\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ These two matrices are already in Jordan canonical form, but since the Jordan blocks are different, they cannot be similar. These matrices both have nilpotence index $2$.
Generically, since nilpotent matrices can only have zero eigenvalues, $\rho(A) = \rho(B)$ and $\rho(A^2)=\rho(B^2)$ is trivially true. The nilpotence index is determined by the largest block in Jordan canonical form. Thus as long as the largest Jordan block in the JCFs of $A$ and $B$ are the same, they can satisfy all the conditions you gave and still not be similar.