Finding eigenvalues of linear transformation given by matrix conjugation 
Let $V = M_n(\mathbb{C})$, and allow $A \in \text{GL}_n(\mathbb{C})$ be an invertible matrix. Define the linear transformation $$C_A: V \to V $$ by $M \mapsto AMA^{-1}$. Find the eigenvalues of $C_A$.

The case is simple if $A$ is diagonizable. Since the inverse of any diagonal matrix is the reciprocal of the entries on the diagonal.
Assume $A$ is diagonizable. If we let $A = SDS^{-1}$, then we have $C_A(M) = C_{SDS^{-1}}(M) = (SDS^{-1})M(SDS^{-1})^{-1} = SDS^{-1}MSD^{-1}S = C_S(C_D(S^{-1}MS))$. Apply the fact about diagonal matrices to get see that this is equal to $C_S(S^{-1}MS) = SS^{-1}MSS^{-1} = M$.
So the only eigenvalue will be $\lambda = 1$.
How do answer this question for any invertible matrix? Since not every invertible matrix is diagonizable, it seems difficult.
So my attempt did not work. Any hints on how to continue?
 A: 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 $A\in GL_n(\mathbb{C})$.

A: 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 $\lambda$.
