# Matrix equation of Kronecker product

$$\ I ⊗ I ⊗ A = Z ⊗ I$$

$$A$$ is a known matrix. $$I$$ is the identity matrix. $$A$$ and $$I$$ are n by n matrices. $$⊗$$ is the Kronecker product. Is there a way to find $$Z$$ in terms of $$A$$ and $$I$$?

A general approach how you can solve equations involving Kronecker products for one of its factors is vectorization: since $$A \otimes B$$ is linear in $$A$$ for given $$B$$ and vice versa, you can rewrite it as $${\rm vec}\{A \otimes B\} = \tilde{B} \cdot {\rm vec}\{A\} = \tilde{A} \cdot {\rm vec}\{B\}$$. This would allow you to solve for the vectorized version of your desired matrix in standard ways (e.g., using Least Squares).
Whether your equation even has a solution will of course depend on the column space of the matrix $$\tilde{A}$$ or $$\tilde{B}$$, respectively.
That said, in your particular problem you are trying to find $$Z$$ such that $$I \otimes A = Z \otimes I$$ (note that the first two identity matrices can be combined into one since the Kronecker product is associative and $$I \otimes I = I$$). Solving this will be tough: $$I \otimes A$$ is a block-diagonal matrix with $$A$$ appearing as repeated blocks. $$Z \otimes I$$ is a matrix containing identity matrices stacked horizontally and vertically and scaled by the elements of $$Z$$. I would claim that this equation only has a solution when $$Z$$ and $$A$$ are diagonal. In this case, you don't need the matrices and their vectorizations, just extract the diagonals and equate those.