<s>[INST] A magic basis of $\mathbb{C}^5$ This is a a small $n$ restriction of another question.
Find a $5\times 5$ matrix of unit vectors $\xi_{ij}\in \mathbb{C}^5$ such that:

*

*Entries along rows and columns are orthogonal, that is for $1\leq i,j,k,l\leq 5$:
$$\delta_{i,k}+\delta_{j,l}=1\implies \langle \xi_{ij},\xi_{kl}\rangle=0.$$

*Entries not on a common row or column are neither parallel, anti-parallel, nor orthogonal, again for $1\leq i,j,k,l\leq 5$::
$$\delta_{i,k}+\delta_{j,l}=0\implies 0<|\langle \xi_{ij},\xi_{kl}\rangle|<1$$
This is proving a real wicked problem. Motivation and an $n=4$ example in the original questions.
Attempts thus far:
I have started with the first two rows :
$$\xi=\frac12\begin{bmatrix} \begin{pmatrix} 2 \\ 0 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix}  & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix}  & \begin{pmatrix} 0 \\ \star \\ \star \\ \star\\ \star\end{pmatrix}
\\  \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix}  & \begin{pmatrix} \star \\ 0 \\ \star \\ \star\\ \star\end{pmatrix}  & \begin{pmatrix} \star  \\ 0 \\ \star \\ \star\\ \star\end{pmatrix} \end{bmatrix}$$
To get orthogonality along the first row, we are multiplying four numbers, and along columns, just three. I started with 24th roots of unity but then this reduced to sixth roots, powers of $w=\exp(2\pi i/6)$. You can get some of what you want with this by considering what combinations of four and three sixth roots give zero. It seems possible to get all but one of:

*

*orthogonal along rows one, two, three

*appropriate orthogonality between rows one and two, and one and three

*appropriate orthogonality between rows two and three

But it seems to fail before it ever gets to rows four or five, or indeed the second condition of being non-orthogonal nor parallel, nor anti-parallel. This wasn't the best I did, but something like:
$$\xi=\frac12\begin{bmatrix} \begin{pmatrix} 2 \\ 0 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1\\ 1\end{pmatrix} & \begin{pmatrix} 0 \\ 1 \\ 1 \\ -1\\ -1\end{pmatrix}  & \begin{pmatrix} 0 \\ w \\ w^4 \\ 1\\ -1\end{pmatrix}  & \begin{pmatrix} 0 \\ 1\\ -1 \\ w^2\\ w^5\end{pmatrix}
\\  \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} w \\ 0 \\ 1 \\ w^2\\ w^4\end{pmatrix} & \begin{pmatrix} w \\ 0 \\ 1 \\ w^5\\ 1\end{pmatrix}  & \begin{pmatrix} w^2 \\ 0 \\ w^4 \\ w^2\\ w\end{pmatrix}  & \begin{pmatrix} w \\ 0 \\ -1 \\ w^4\\ -1\end{pmatrix}
\\  \begin{pmatrix} 0 \\ 0 \\ 2 \\ 0\\ 0\end{pmatrix} & \begin{pmatrix} 1\\ w \\ 0 \\ w^5\\ -1\end{pmatrix} & \begin{pmatrix} w^2 \\ w \\ 0 \\ w^2\\ 1\end{pmatrix}  & \begin{pmatrix} -1\\ w^2 \\ 0 \\ w^5\\ 1\end{pmatrix}  & \begin{pmatrix} w^2 \\ w \\ 0 \\ w\\ w^2\end{pmatrix}  \end{bmatrix}$$
This attempt falls down on orthogonality in row three, and orthogonality in row 2 vs row 3.
 [/INST] David Roberson has pointed me towards Figure 1 from Quantum symmetry vs nonlocal symmetry.
This is a so-called quantum latin square:

The produced quantum Latin squares are indexed by group elements and have the property that the inner product of the $i,j$-entry and $k,l$-entry depends only on $i^{-1}k$ and $j^{-1}l$. I would not be surprised if one can find a quantum Latin square meeting your requirements for all $n \geq 5$ by using this construction on the cyclic group $\mathbb{Z}_n$ (and picking the right permutation of its characters).

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