Fourier matrix

The Fourier matrix of size $N$ reads $$\left[F^{'}_N\right]_{j,k}=\frac{1}{\sqrt{N}}\exp\left(2\pi i(j-1)(k-1)\frac{1}{N}\right):j,k=1,2,...,N$$ and is unitary. Hence the matrix $F_N=\sqrt{N}F^{'}_N$ is a complex Hadamard matrix in a dephased form.

Equivalence between Kronecker (tensor) products of Fourier matrices.

A Kronecker product of Fourier matrices $k_1=F_{m(1)}\otimes \cdots \otimes F_{m(p)}$ is equivalent (even permutation equivalent) to any other such product of the same total size $k_2=F_{n(1)}\otimes \cdots \otimes F_{n(r)}$ if and only if $k_2$ can be obtained from $k_1$ by a combination of an arbitrary number of operations from the set:

changing the order of Kronecker product factors,
replacing a subproduct $F_{m(k)} \otimes F_{m(k+1)}$ by the factor $F_{m(k) \cdot m(k+1)}$ only if $m(k)$ and $m(k+1)$ are relatively prime,
replacing a factor $F_{ab}$ by the subproduct $F_a \otimes F_b$ only if $a$ and $b$ are relatively prime [28].

Example 1

$F_2\otimes F_3\simeq F_6$
$F_2\otimes F_2\otimes F_{15}\simeq F_2\otimes F_{30}$
$F_4\otimes F_6\not\simeq F_{24}$
$F_8\otimes F_8\not\simeq F_{64}$
$F_8\otimes F_4\not\simeq F_8\otimes F_2\otimes F_2$ the same types of elements on both sides!
$F_8\otimes F_6\otimes F_4\not\simeq F_{48}\otimes F_4$

Example 2

Equivalence classes of Kronecker products of Fourier matrices, of the total size $N = 72 = 3 \cdot 3 \cdot 2 \cdot 2 \cdot 2$. The elements in each class $[ x ]$, represented by $x$, are listed up to the order of Kronecker product factors:

$\left[F_9\otimes F_8\right]=\{F_9\otimes F_8, F_{72}\};$ reorderings
$\left[F_9\otimes F_4\otimes F_2\right]= \{F_9\otimes F_4\otimes F_2, F_{18}\otimes F_4, F_{36}\otimes F_2\};$ reorderings
$\left[F_9\otimes F_2\otimes F_2\otimes F_2\right]= \{F_9\otimes F_2\otimes F_2\otimes F_2, F_{18}\otimes F_2\otimes F_2\}; $ reorderings
$\left[F_3\otimes F_3\otimes F_8\right]= \{F_3\otimes F_3\otimes F_8, F_{24}\otimes F_3\}; $ reorderings
$\left[F_3\otimes F_3\otimes F_4\otimes F_2\right]= \{F_3\otimes F_3\otimes F_4\otimes F_2, F_6\otimes F_3\otimes F_4, F_{12}\otimes F_3\otimes F_2, F_{12}\otimes F_6\}; $ reorderings
$\left[F_3\otimes F_3\otimes F_2\otimes F_2\otimes F_2\right]= \{F_3\otimes F_3\otimes F_2\otimes F_2\otimes F_2, F_6\otimes F_3\otimes F_2\otimes F_2, F_6\otimes F_6\otimes F_2\}; $ reorderings