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This question is linked to the discrete Fourier transform. If $u=[1,\cdots,1]^T$, then $A=\begin{pmatrix}0&u^T\\u&B\end{pmatrix}$ and the equation $AA^T=(n-1)I_n$ is equivalent to $Bu=0,BB^T=(n-1)I_{n-1}-uu^T$. The spectrum of $BB^T$ is $0$ and $n-1$ ($n-2$ times). Let $F$ be the $(n-2)\times(n-2)$ matrix defined by $f_{j,k}=\exp(\dfrac{2i\pi j(n-k-1)}{n-1})$ (the Fourier matrix associated to the non-zero eigenvalues of $B$). $B,B^T$ commute and they are simultaneously diagonalizable (using the complete Fourier matrix) to $B'=diag(F[x_1,\cdots,x_{n-2}]^T)$ and to ${B'}^T=diag(F[x_{n-2},\cdots,x_1]^T)$. The eigenvalues of $BB^T$ are the $B'_{i,i}{B'}^T_{i,i}$. Finally our vector $X$ is solution of the quadratic system: for every $i\leq n-2$, $(\sum_{j=1}^{n-2}f_{i,j}x_j)(\sum_{j=1}^{n-2}f_{i,j}x_{n-1-j})=n-1$.