Let $A$ be an arbitrary diagonalizable square matrix. Consider $A \oplus A = A \otimes I + I \otimes A$ that is a Kronecker sum of $A$ with itself ($I$ is of the same size as $A$ here). I want to express the eigenvectors of $A \oplus A$ through the eigenvectors of $A$ explicitly.
If $A = S_A \cdot D_A \cdot S_A^{-1} $ (eigendecomposition), then the set of right eigenvectors of $A \otimes I$ is, obviously, $S_A \otimes I$, while for $I \otimes A$ it is $I \otimes S_A$. What I also know is that $A \otimes I$, and $I \otimes A$ commute, therefore, they are simultaneously diagonalizable.
How can one express the matrix $S_{A \oplus A}$ that diagonalizes the corresponding matrix through $S_A$?
Take $S_{A \oplus A} = S_A \otimes S_A$. Indeed, we have $$ S_{A \oplus A}^{-1}(A \oplus A)S_{A \oplus A} =\\ (S_A \otimes S_A)^{-1}(A \otimes I + I \otimes A)(S_A \otimes S_A) =\\ (S_A^{-1} \otimes S_A^{-1})(A \otimes I + I \otimes A)(S_A \otimes S_A) =\\ (S_A^{-1} \otimes S_A^{-1})(A \otimes I)(S_A \otimes S_A) + (S_A^{-1} \otimes S_A^{-1})(I \otimes A)(S_A \otimes S_A) =\\ (S_A^{-1}AS_A) \otimes (S_A^{-1}IS_A) + (S_A^{-1}IS_A) \otimes (S_A^{-1}AS_A) =\\ D_A \otimes I + I \otimes D_A. $$ I will leave it to you to verify that $D_{A \oplus A}:= D_A \otimes I + I \otimes D_A$ is indeed a diagonal matrix.