Is a normal matrix satisfying $A^TA=...$ circulant? 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.
 A: For matrices of order $3$ one deals with three vectors, each in the positive quadrant of a coordinate plane, in the sphere of radius $\sqrt{a}$. The fact that all escalar products are equal (equal to $b$) says that these three vectors form a spherical equilateral triangle. Now, such a triangle with vertices sitting in the coordinate planes is completely determined by one single vertex. This follows because the sides of the triangles are maximum circles on the sphere (one also can carefully write a system of equations and solve them). In any case, since there is only one triangle as wanted, the obvious one (circulating the coordinates) must be it. It is worth to remark that one doesn't need here the matrix to be normal, just the condicion on the product $A^TA$. 
Furthermore, after some elementary discussion, one sees that given the parameters $a,b>0$, there is a $3\times 3$ matrix $A$ with $A^TA$ as wanted if and only if $a>2b$ and then the matrix is circulant and normal.
This approach breaks down in higher dimensions. The matrix
$$
\begin{pmatrix}
0&1/3&1/2&1/4\\ 1/3&0&1/4&1/2\\ 1/4&1/2&0&1/3\\ 1/2&1/4&1/3&0
\end{pmatrix}
$$
verifies the conditions with $a=61/144$ and $b=1/4$ (hope there's no silly mistake). Such an example comes from a lot of computations to understand the issue. Namely:
1) Look for all circulant matrices of order $4$ that verify the hypotheses.
First one can suppose $a=1$ by scaling the whole matrix. Then one has some system in three indeterminates, but ends up with a certain degree $4$ polynomial that must have a positive root. Just for the record my computations give
$$
P(t)=16t^4+16(b-1)t^3+8b(3b-2)t^2+4b^2(b-1)t+b^4=0.
$$
Hence given $b$ there are very few ($\le 4$) circulant matrices as wanted. 
2) Look for non-circulant solutions. Since dropping the circulant condition the number of indeterminates in the problem passes from 3 to 12, there seems to be more room for solutions to the latter problem. Actually one can produce a monster like this
$$
\begin{pmatrix}
0&\frac{2b(x+y)}{2b+(x+y)^2}&\frac{x+y}{2}&\frac{2b-x^2+y^2}{2(x+y)}\\ 
\frac{2b(x+y)}{2b+(x+y)^2}&0&\frac{2b+x^2-y^2}{2(x+y)}&\frac{x+y}{2}\\ 
\frac{b}{x+y}&x&0&\frac{2b(x+y)}{2b+(x+y)^2}\\ 
y&\frac{b}{x+y}&\frac{2b(x+y)}{2b+(x+y)^2}&0
\end{pmatrix}.
$$
It verifies the $A^TA$ condition off the diagonal (for $b$) and it remains to impose the columns and the rows to have equal norm. This in the end gives $x=y$. Also, one sees that the condition for the matrix to be circulant is $b=2x^2$ and then all entries off the diagonal are $\equiv x$. 
Here I just short cut and look for some concrete example as above ($x=y$ with $b\ne2x^2$). 
In any case, even if my computations fail, the problem in dimension $4$ should be tackled by explicit solutions of quadratic systems.
Hope this gives some insight. For instance, all computations above suggest that although not circulant, all columns must be permutations of the first.
