Here's an example of $A . B = scalar \times C$, done with magic squares. The last square does not have a consecutive range of digits.
Drop the magic square requirement. In $2\times2$ matrices we have the following, where a matrix times a scalar is factored into permutations of the original matrix. All entries of the matrix are distinct.
$$(\begin{pmatrix} -1 & 0 \\ 2 & 1 \\ \end{pmatrix} + a)\cdot (\begin{pmatrix} 1 & 0 \\ -1 & 2 \\ \end{pmatrix}+a) = (2 a + 1)\times(\begin{pmatrix} -1 & 0 \\ 1 & 2 \\ \end{pmatrix}+a)$$
Here are examples with $3 \times 3$ matrices. Can these be canonicalized in some way and all solutions listed?
$$ \begin{pmatrix} 3 & 2 & 4 \\ 1 & -3 & -1 \\ -4 & 0 & -2 \\ \end{pmatrix} \cdot \begin{pmatrix} -3 & -1 & 2 \\ -2 & 0 & 3 \\ 4 & 1 & -4 \\ \end{pmatrix} = \begin{pmatrix} 3 & 1 & -4 \\ -1 & -2 & -3 \\ 4 & 2 & 0 \\ \end{pmatrix}$$
$$ \begin{pmatrix} 4 & -2 & 1 \\ -3 & 2 & -1 \\ -4 & 3 & 0 \\ \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 & -2 \\ 2 & 0 & -4 \\ 3 & 4 & -3 \\ \end{pmatrix} = \begin{pmatrix} 3 & 0 & -3 \\ -2 & -1 & 1 \\ 2 & 4 & -4 \\ \end{pmatrix}$$
$$ \begin{pmatrix} -1 & 0 & 1 \\ 3 & 4 & -3 \\ -2 & -4 & 2 \\ \end{pmatrix} \cdot \begin{pmatrix} 4 & -1 & -3 \\ 0 & -2 & 2 \\ 3 & -4 & 1 \\ \end{pmatrix} = \begin{pmatrix} -1 & -3 & 4 \\ 3 & 1 & -4 \\ -2 & 2 & 0 \\ \end{pmatrix}$$
$$ \begin{pmatrix} 4 & -3 & -4 \\ -1 & 1 & 0 \\ -2 & 2 & 3 \\ \end{pmatrix} \cdot \begin{pmatrix} -1 & 4 & -2 \\ 2 & 0 & -4 \\ -3 & 3 & 1 \\ \end{pmatrix} = \begin{pmatrix} 2 & 4 & 0 \\ 3 & -4 & -2 \\ -3 & 1 & -1 \\ \end{pmatrix}$$
Is there a way to solve "Factor $A =\begin{pmatrix} -4 & 0 & -3 \\ 3 & 4 & -1 \\ 1 & 2 & -2 \\ \end{pmatrix}$ into permutations of $A$." ?
Can examples be made with larger matrices?
Try $$ \pmatrix{7 & -3 & 6 & 8 \cr 1 & -5 & 4 & -1 \cr -7 & 3 & -4 & -6 \cr 0 & -2 & 5 & 2 \cr} \pmatrix{ -5 & 3 & 7 & -1 \cr -6 & 4 & 8 & -2 \cr -4 & 2 & 6 & 0 \cr 5 & -3 & -7 & 1 \cr} =\pmatrix{ -1 & -3 & 5 & 7 \cr 4 & -6 & -2 & 8 \cr 3 & 1 & -7 & -5 \cr 2 & -4 & 0 & 6 \cr}$$ and $$ \pmatrix{ 4 & -6 & 0 & 8 \cr -2 & -1 & 3 & 1 \cr -5 & -4 & -7 & -3 \cr 7 & 6 & 5 & 2 \cr} \pmatrix{5 & -4 & -6 & -2 \cr -5 & 4 & 7 & 1 \cr 3 & -1 & -3 & 0 \cr -7 & 6 & 8 & 2 \cr} =\pmatrix{ -6 & 8 & -2 & 2 \cr -3 & 7 & 4 & 5 \cr -5 & -7 & -1 & 0 \cr 6 & 3 & 1 & -4 \cr} $$