Finding eigenvalues of linear transformation given by matrix conjugation

You can make a continuity argument to reduce to the case of diagonalizable matrices. The characteristic polynomial of $C_A$ varies continuously with $A$, and diagonalizable matrices are dense in $GL_n(\mathbb{C})$ (for instance, because every matrix is conjugate to an upper triangular one, and an upper triangular matrix can always be perturbed to a diagonalizable one by just making the diagonal entries distinct).

So, if you know the eigenvalues (with their multiplicities) of $C_A$ when $A$ is diagonal, you can deduce them for arbitrary $A$ by continuity. In the case that $C_A$ is diagonal, you can write down what $C_A$ does to the entries of a matrix quite explicitly to find its eigenvalues.

More details on how to finish are hidden below.

Suppose $A$ is diagonalizable with diagonal entries $a_1,\dots,a_n$. Then $C_A$ multiplies the $ij$ entry of a matrix by $a_i^{-1}a_j$ (since left multiplication by $A$ multiplies the $j$th column by $a_j$ and right multiplication by $A^{-1}$ multiplies the $i$th row by $a_i^{-1}$). In other words, with respect to the standard basis on $M_n(\mathbb{C})$, $C_A$ is diagonal with diagonal entries $a_i^{-1}a_j$.

Thus, if $A$ is any diagonalizable matrix, the eigenvalues of $C_A$ (with multiplicity) are $a_i^{-1}a_j$, where the $a_i$ are the eigenvalues of $A$. It follows by continuity that the same is true for arbitrary $A\in GL_n(\mathbb{C})$.

This post about which matrices commute up to a scalar may be helpful, as you are looking for $M$ and $\lambda$ such that $AMA^{-1} = \lambda M$, essentially requiring $M$ and $A$ to change places and leave behind a scalar $\lambda$.