TITLE: Is a normal matrix satisfying $A^TA=...$ circulant?
QUESTION [6 upvotes]: Let $A=\{a_{ij}\}$ be a normal matrix such that $a_{ij}\geq 0$ with equality iff $i=j$. Suppose that
$$
A^TA=\begin{pmatrix}
a & b & \cdots & b\\
b & a & \ddots & \vdots\\
\vdots & \ddots & a & b\\
b & \cdots & b & a\\
\end{pmatrix},\ where\ b>0.
$$
Does it follow that $A$ is a circulant matrix?
Note: There is a partial classification of non-negative normal matrices posted here, which seems like it can be used to attack this problem.
There is a geometric interpretation as well: both the set of rows and the set of columns of $A$ form equidistant sets of vectors on a sphere, and basic geometry appears to severely restrict the possibilities.

Reposted from math.SE

REPLY [5 votes]: NO.
Let $A$ satisfy the assumptions. If $P$ is a permutation matrix, then $B:=PA$ satisfies the assumptions too: on the one hand, we have $B^TB=A^TP^TPA=A^TA$. On the other hand (remark that the matrix in the question is permutation-invariant)
$$BB^T=PAA^TP^T=PA^TAP^T=A^TA=B^TB.$$
If the claim is true, we find that $B=PA$ is circulant for every  permutation matrix $P$. We deduce that $a_{ij}$ not only depends upon $j-i$ (modulo $n$), but also depends only upon $j-\sigma(i)$, for every $\sigma\in{\frak S}_n$. We conclude that $a_{ij}$ is constant.
Therefore every matrix $A$ satisfying the assumptions, such that $a_{ij}$ is not constant, provides a counter-example of the form $PA$ for some (many) permutation matrices $P$.