Let $\Sigma_1, \Sigma_2\in\mathbb{R}^{d\times d}$ symmetric and positive definit and $N\in\mathbb{N}$, $N\neq d$.
Can the following term be simplified?
$$tr((I_N\otimes \Sigma_2)(\Sigma_1\otimes I_N))$$
where tr is the trace, $I_N$ is the identity in $\mathbb{R}^N$ and $\otimes$ is the Kronecker symbol.
$ \def\a{\alpha}\def\b{\beta} \def\A{{\cal A}}\def\B{{\cal B}} \def\LR#1{\left(#1\right)} \def\BR#1{\Bigl(#1\Bigr)} \def\T{\operatorname{Tr}} \def\v{\operatorname{vec}} $Define the variables $$\eqalign{ I_n &\in {\mathbb R}^{n\times n}\qquad &\quad\; e_n &= \v(I_n) \\ K &\in {\mathbb R}^{nd\times nd}\quad &K^TK &= I_{nd} \\ L &= \LR{I_d\otimes K\otimes I_n}\quad &L^TL &= I_{n^2d^2} \\ a &= \v(\Sigma_1) \quad &\quad\;\;b &= \v(\Sigma_2) \\ \A &= \LR{\Sigma_1\otimes I_n} = \A^T\qquad &\quad\;\;\a &= \v(\A) = L\LR{I_{d^2}\otimes e_n}a \\ \B &= \LR{I_n\otimes\Sigma_2} = \B^T \qquad &\quad\;\;\b &= \v(\B) = L\LR{e_n\otimes I_{d^2}}b \\ }$$ where $I_n$ is the identity matrix and $K=K^{(n,d)}$ is the commutation matrix
Use these variables to expand the expression for the trace $$\eqalign{ \T\LR{\A^T\B} &= \a^T\b \\ &= \BR{L\LR{I_{d^2}\otimes e_n}a}^T\BR{L\LR{e_n\otimes I_{d^2}}b} \\ &= a^T {\BR{I_{d^2}\otimes e_n^T}\BR{e_n\otimes I_{d^2}}}\,b \\ &= a^TMb \\ &= \T\LR{Mba^T} \\ }$$