$A$ similar to $B$ implies $A\otimes 1 - 1 \otimes A$ is similar to $A\otimes 1 - 1\otimes B$

Question

Let $A$ and $B$ be similar linear transformations on a finite dimensional vector space $V$ over an algebraically closed field. I.e. there exists an invertible transformation $C$ such that $B=C^{-1}AC$. How one can show that then the transformations $A\otimes 1 - 1\otimes A$ and $A\otimes 1-1\otimes B$ are also similar to each other viewed as transformations of $V\otimes V$? Here $1$ is the identity transformation on $V$. The authors of the article I am reading claim that this is very easy.

Answer 1

$(1\otimes C)^{-1}(A\otimes 1 - 1\otimes A)(1\otimes C) = A\otimes 1 - 1\otimes B$