Well, one thing is special about it, but it takes a while to explain.

**Please let me know, whether this number occurs in other special occasions as well.**

The explanation: Let $p$ be a complex polynomial of degree $n$ with simple zeros $z_j$ and simple critical points $z'_k$ and consider the matrix

\begin{equation} A_p \ := \ \begin{pmatrix} \frac{1}{n} & \ldots & \frac{1}{n} \\ \frac{1}{n} \, \frac{H_c(z'_1)}{(z_1 - z'_1)^2} & \ldots & \frac{1}{n} \, \frac{H_c(z'_1)}{(z_n - z'_1)^2} \\ \vdots & & \vdots \\ \frac{1}{n} \, \frac{H_c(z'_{n-1})}{(z_1 - z'_{n-1})^2} & \ldots & \frac{1}{n} \, \frac{H_c(z'_{n-1})}{(z_n - z'_{n-1})^2} \\ \end{pmatrix} \, , \end{equation} where $H_c(z'_k)$ is defined as follows: $$ H_c(z'_k) \ := \ \frac{n}{\sum^{n}_{j=1} \frac{1}{(z_j - z'_k)^2}} \, . $$

It is a complex relative gain array, so its row and column sums equal 1 and it has eigenvalues 0 and 1, shared with $A^t_pA_p$, $A_pA^t_p$, $A^*_pA_p$, and $A_pA^*_p$, the former complex symmetric, the latter hermitian.

Some years ago I proved $A^t_pA_p$ also has the eigenvalue $n-2/n$ and the corresponding eigenvector $(z_1, \ldots, z_n)^t$, if we assume the centroid of the zeros to be zero itself. $A_pA^t_p$ has the corresponding eigenvector $(0, z'_1, \ldots, z'_{n-1})^t$, whose centroid is zero too in that case.

If this eigenvalue is simple, then the matrix governs the geometry of the zeros of $p$, as there will be a euclidean transformation/scaling transforming any set of zeros generating $A_p$ to any other set generating the same $A_p$, $A_p$ is invariant concerning these transformations.

Unfortunately, this eigenvalue is not always simple, for example consider $A_q$ of $q(z) \, = \, z^4 - z$.

That was my knowledge of this particular situation until some pupils at my school considered rhombi and rectangles, whose vertices were taken as zeros of the considered polynomials, under my guidance.

We found the following:

Let $r(z) \, = \, (z^2 - r^2)(z^2 + 1)$, where $1 \, < \, r \, < \, \infty$. Then $A_rA^t_r \, = \, A_qA^t_q$ if and only if $r \, = \, 2 + \sqrt{3}$, corresponding to a double eigenvalue 1/2 indeed.

So there are some different geometric configurations leading to the same matrices but they are very rare.

Ever since then I have asked myself the question given as headline. I have no idea where to look and would be glad for any suggestions. It might well be just an isolated curiosity, but this is the place to find out, I assume.