Counting symmetric unitary matrices with elements of equal magnitude Let $X$ be an $n\times n$ symmetric unitary matrix with elements of equal magnitude and the elements of the first row (and the first column, of course) are $1/\sqrt{n}$, i.e. $X_{j,k} = e^{i \phi_{j,k}}/\sqrt{n}$ with $\phi_{j,k}=\phi_{k,j} \in \mathbb{R}$ and $\phi_{1,k}=0$.
Prove (or disprove) that such matrix is unique up to permutations.
The background here is to get understanding of the set of bases in $\mathbb{C}^n$ with coordinates differ by phase factors only. The orthogonality relation $(v_j, v_k) = n^{-1}\sum_l \exp(i(\phi_{j,l}-\phi_{k,l})) = 0$ for $j \ne k$ is invariant with respect to $\phi_{j,k} \to \phi_{j,k}+\omega_j + \omega_k$. This would be great if the set could be obtained from a particular choice
$$X_{j,k} = \frac{1}{\sqrt{n}}\exp\left(\frac{2\pi i}{n} (j-1)(k-1)\right)$$
by permutations and the symmetry transformation.
The statement seems plausible since the matrix $X$ is determined by $n(n-1)/2$ phases with $n(n-1)/2$ constraints imposed by orthogonality relations. The proof, however, is lacking. I would greatly appreciate help in a form of proof/disproof or a strong hint.
 A: The matrix in question is not unique, at least not in all dimensions.
Consider your chosen matrix $X$ with $n=4$; this is the discrete Fourier matrix
$$X=\frac{1}{2}\begin{pmatrix}1&1&1&1\\
1&i&-1&-i\\
1&-1&1&-1\\
1&-i&-1&i
\end{pmatrix}$$
Another matrix with $n=4$ that satisfies your criteria is the Hadamard matrix
$$H=\frac{1}{2}\begin{pmatrix}1&1&1&1\\
1&1&-1&-1\\
1&-1&-1&1\\
1&-1&1&-1
\end{pmatrix}$$
Because the two matrices share the same first row and column, they cannot be obtained from each other by multiplying rows or columns by phases.
Now, your statement is true when $n=1$ (trivially), $n=2$ (because the condition on the first row and column leaves only a single element to be determined), and $n=3$ (where the two possible matrices are equivalent up to a permutation of the final two rows or columns). The degree of freedom argument that you presented breaks down because all of the phases in $X$ depend on the same root of unity. So, one could still ask whether there are other dimensions for which your statement is true. For example, is it true for prime dimensions?
