In transforming matrices between bases I have come across this curious equation: $\mathbf{A} \mathbf{B} \mathbf{C} = \mathbf{A}^{\mathsf{T}} \mathbf{B} \mathbf{C}^{\mathsf{T}}$, for all $\mathbf{B}$. I have tried massaging this equation every which way and have failed to relate it to a well-known matrix property. It seemed from the context in which it arose (I was relating the transpose in of a matrix to the transpose in another set of bases) that it might have something to do with orthogonality, and I have tested this property with orthogonal matrices and it seems to hold, but I cannot derive orthogonality from it. I have noticed that if $\mathbf{A}$ and $\mathbf{C}$ are orthogonal an equivalent form would be that $\mathbf{A}^2 \mathbf{B} \mathbf{C}^2 = \mathbf{B}$, but this does not help in finding the meaning of this equation.
Is this property equivalent to another property, and if not, does it have a name?
Let us work with Einstein convention (sum over repeated indices) to rewrite the condition $ABC=A^tBC^t$ as $$ A_{ij}B_{jk}C_{kl}=A_{ji}B_{jk}C_{lk},\qquad \forall i,l. $$ Take $B$ to be an elementary unit matrix, namely $B_{jk}=\delta_{jp}\delta_{kq}$ for some $p,q$, to get the explicit form of the relation between $A,C$ in terms of their entries: $$ A_{ip}C_{ql}=A_{pi}C_{lq},\qquad \forall i,l,p,q. $$ This relation might be interpreted intrinsically by looking at the tensor map $A\otimes C$ (aka Kronecker product), then it simply says $$ A\otimes C = A^t \otimes C^ t = (A\otimes C)^ t, $$ namely, $A\otimes C$ is symmetric.