Matrix identity involving Kronecker product It is well-known that, given real matrices $\mathbf{A}_1$, $\mathbf{A}_2$ and $\mathbf{B}$ with dimensions $m_1 \times n_1$, $m_2 \times n_2$ and $n_1 \times n_2$, respectively, then
$$(\mathbf{A}_2 \otimes\mathbf{A}_1)\text{vec}(\mathbf{B}) = \text{vec}(\mathbf{A}_1\mathbf{B}\mathbf{A}_2^\top),$$
where $\text{vec}$ denotes the vectorization operator, $\otimes$ is the Kronecker product and $^\top$ is the transposition operation.
I wonder that similar relations can be deduced when more matrices $\mathbf{A}_i$, $i=1,\ldots, m$, are involved (and hence, $\mathbf{B}$ is a tensor with order $m$), that is to say, how can we express the product
$$(\mathbf{A}_m \otimes \cdots \otimes\mathbf{A}_1)\text{vec}(\mathbf{B}) = \quad ?$$
Any help would be appreciated.
 A: $\def\v{\operatorname{vec}}\def\o{\otimes}\def\B{\,\color{red}{B}\,}$By
reshaping $\v(B)$ we can create a matrix $\B$
Consider the case for $m=4$
$$\eqalign{
\v\Big((A_1) \B (A_4^T\o A_3^T\o A_2^T)\Big)
 &= (A_4\o A_3\o A_2\o A_1)\v(B) \\
\v\Big((A_2\o A_1) \B(A_4^T\o A_3^T)\Big)
 &= (A_4\o A_3\o A_2\o A_1)\v(B) \\
\v\Big((A_3\o A_2\o A_1) \B(A_4^T)\Big)
 &= (A_4\o A_3\o A_2\o A_1)\v(B) \\
}$$
Extrapolating, if the number of $A$ matrices is $m$, then there are $(m-1)$ locations where the $\color{red}{B}$ matrix can be placed, which will yield the same result upon vectorization.
The $A$ matrices are cyclically permuted to accommodate the shifting $\color{red}{B}$ matrix.
Another approach is calculate the Kronecker factorization of the vector
into constituent vectors whose dimensions are compatible with the $A$ matrices
$$\eqalign{
&\v(B) = b_4\o b_3\o b_2\o b_1 \\
&(A_4\o A_3\o A_2\o A_1)\v(B) = (A_4b_4\o A_3b_3\o A_2b_2\o A_1b_1) \\
}$$
Although it may not be possible to find a monomial factor, it is always possible to find a sum of such factors, e.g.
$$\eqalign{
\v(B) &= \sum_{k=1}^K b_{4k}\o b_{3k}\o b_{2k}\o b_{1k} \\
}$$
