Consider an orthonormal matrix $W\in\mathbb{R}^{2n\times 2n}$ that satisfies the "abs property": $|w_i|^T |w_{i+n}|=1$$$|w_i|^T |w_{i+n}|=1$$ for all $i=\{1,2,\ldots,n\}$. Here$i \in \{1,2,\ldots,n\}$, where $w_i\in\mathbb{R}^{2n}$$w_i \in \mathbb{R}^{2n}$ is the $i$-th column of $W$ and $|w_i|$ denotes the vector of absolute values of $w_i$.
It can be shown that $(|w_i| - |w_{i+n})^T(|w_i| - |w_{i+n}|)=0$, so $|w_i| = |w_{i+n}|$, for all $i=\{1,2,\ldots,n\}$. In other words, the last $n$ columns are the first $n$ columns with a potential switch of the sign in each entry.
Feels like $W$ is a permutation of some $$G=\begin{pmatrix}U/\sqrt{2} & U/\sqrt{2}\\\ V/\sqrt{2} & -V/\sqrt{2}\end{pmatrix}$$
$$G = \frac{1}{\sqrt{2}} \begin{pmatrix} U & U \\ V & -V \end{pmatrix}$$
(more exactly $W=PG$ for some permutation $P$), where $U$ and $V$ are orthonormal matrices from $\mathbb{R}^{n\times n}$.
Obviously, $PG$ is orthonormal and satisfies abs property. Is the reverse true? Can all orthonormal $W$s with the abs property be written as left/row permutations of $G$? Maybe this is obvious to folks knowledgeable of orthogonal groups? I was not able to prove it just with elementary linear algebra.
Also, does the abs property of $W$ or the structure of $G$ rings a bell in terms of terminology? I tried generalized or block rotation matrices, but nothing showed up.
