On composition with vec operation Given matrix $$A=CDE$$ we have $$vec(A)=(E'\otimes C)vec(D)$$
Does $$A=BCDEF$$ give $$vec(A)=(F'\otimes B)(E'\otimes C)vec(D)?$$

Yes, it does! One property of the Kronecker (tensor) product which is worth knowing is that
$$
(A \otimes B)(C \otimes D) = (AC) \otimes (BD)
$$
(assuming the dimensions of the matrices are compatible).  So, your expression on the right is 
$$
(F' \otimes B)(E' \otimes C)vec(D) = \\
[(F'E') \otimes (BC)]vec(D) = \\
[(EF)' \otimes (BC)]vec(D)
$$
which is simply an application of your first formula.
