I have a question regarding eigenvalues of complex matrices. Let's assume that $Z \in \mathbb{C}^{n \times n}$ with eigenpair $(\lambda_i,\gamma_i)$. We now look at
$B\gamma_i^*=\lambda_i^* \gamma_i^* \tag{1}$
where $(\bullet)^*$ denotes complex conjugate. If $Z$ is diagonalizable, then $Z$ has a unique set of eigenpairs, so in this case we know that $B=Z^*$ (please correct me if I'm wrong). My question is: if $Z$ is defective, can I find a matrix $B = Z$ such (1) is valid? In other words, can I form a defective matrix that shares eigenpair set with its own conjugate?
Thanks in advance.