Consider an orthonormal matrix $W\in\mathbb{R}^{2n\times 2n}$ that satisfies the "abs property" $$|w\_i|^T |w\_{i+n}|=1$$ for all $i \in \{1,2,\ldots,n\}$, where $w\_i \in \mathbb{R}^{2n}$ 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


$$G = \frac{1}{\sqrt{2}} \begin{pmatrix} U & U \\ V & -V \end{pmatrix}$$


(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.



Orthonormal matrices with columns that switch signs
This is not true.


For example, consider $H\_{12}$: the $12 \times 12$ [Hadamard matrix](https://en.wikipedia.org/wiki/Hadamard_matrix). Well, $H\_{12}/\sqrt{12}$ is an orthonormal matrix with the "abs property", but the matrices $U$ and $V$ cannot possibly exist since $\sqrt{6}U$ and $\sqrt{6}V$ would be $6 \times 6$ Hadamard matrices (which don't exist).


