Unitary matrix with elements of equal modulus It is a well-known fact that there are unitary matrices $U$, whose elements have equal modulus ($|u_{ij}|=c, \forall i,j$). The most known example is the DFT Matrix.
My question is, are the DFT Matrices and their permutations the only matrices that satisfy this relation? A similar question has been asked in this forum before (alas, with no answer), but with the constraint of symmetry. In my case, I do not care if the matrix is symmetric or not.
 A: Your conjecture is false because you can multiply each column of a solution by any complex of modulus $1$.
When $n=2$, the set of solutions is $V(2)=\{\dfrac{1}{\sqrt{2}}\begin{pmatrix}e^{iu}&e^{iv}\\e^{i(u+\alpha)}&-e^{i(v+\alpha)}\end{pmatrix};u,v,\alpha\in\mathbb{R}\}$.
Note that $V(2)$ is a real variety of dimension $3$ and that $U(2)$ is a real variety of dimension $4$.
EDIT. When $n=3$, we obtain $V(3)=\{\dfrac{1}{\sqrt{3}}\begin{pmatrix}e^{iu_1}&e^{iu_2}&e^{iu_3}\\e^{iu_4}&j^2e^{i(u_4-u_1+u_2)}&je^{i(u_4-u_1+u_3)}\\e^{iu_7}&je^{i(u_7-u_1+u_2)}&j^2e^{i(u_7-u_1+u_3)}\end{pmatrix};u_1,u_2,u_3,u_4,u_7\in\mathbb{R};1+j+j^2=0\}$.
Thus $dim(V(3))=5$ while $dim(U(3))=9$.
EDIT. Answer to @Bryson of Heraclea. Yes, all the possible cases. Yet, beware, the case $n=3$ is special; indeed, when the sum of $3$ complex numbers $a,b,c$ of modulus $1$ is $0$, then $b=ja,c=j^2a$; this result doesn't generalize to higher dimensions;
A: The answer is no. E.g. by halving a Hadamard matrix $H_4$ of order $4$, you get a real orthogonal matrix
$$
\frac12H_4=\pmatrix{
1& 1& 1& 1\\
1&-1& 1&-1\\
1& 1&-1&-1\\
1&-1&-1& 1},
$$
but it is neither a permutation matrix nor a DFT matrix.
