There exists $C\neq0$ with $CA=BC$ iff $A$ and $B$ have a common eigenvalue Question:
Suppose $V$ and $W$ are finite dimensional vector spaces over $\mathbb{C}$. $A$ is a linear transformation on $V$,  $B$ is a linear transformation on $W$. Then there exists a non-zero linear map $C:V\to W$ s.t. $CA=BC$ iff $A$ and $B$ have a same eigenvalue.
===========This is incorrect==========
Clearly, if $CA=BC$, suppose $Ax=\lambda x$, then $B(Cx)=CAx=C(\lambda x)=\lambda (Cx)$, so $A$ and $B$ have same eigenvalue $\lambda$. On the other hand, if $A$ and $B$ have a same eigenvalue $\lambda$, suppose $Ax=\lambda x, By=\lambda y$. Define $C:V\to W$ s.t. $Cx=y$, then $BCx=By=\lambda y=C\lambda x=CAx$. But I don't kow how to make $BC=CA$ for all $x\in V$. 
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Here is a simple solution for the if condition.
Let $\lambda$ be the common eigenvalue. Let $u$ be a right eigenvector for $B$, that is 
$$Bu= \lambda u$$
and $v$ be a left eigenvector for $A$, that is 
$$v^TA= \lambda  v^T$$
Then $C =uv^T$ is a non-zero $n \times n$ matrix which works:
$$CA = u v^T A = \lambda u v^T =\lambda C$$
$$BC= B u v^T= \lambda u v^T= \lambda C$$
