Let $m\ge 3$ be an integer and $r$ be any positive ingeger. Denote $\cal I$ to be a $m$ order $r$ dimensional idnetity tensor; that is, $\cal I_{i_1...i_m} = 1$ if $i_1=...=i_m$ and $0$ otherwise. Suppose there is a matrix $O$ satisfying that $$\cal I \times_1 O \times ...\times_m O = \cal I.$$For any matrix $\alpha$, do we always have $||Vec(\alpha O)||_m = ||Vec(\alpha)||_m$? Here $Vec(\cdot)$ is the vectorization operator and for any $a\in \mathbb{R}^n$, $||a||_m^m = \sum_{i=1}^n|a|^m$.
An equavalent formulation of the formula $\cal I \times_1 O \times ...\times_m O = \cal I$ is that for any $i_1, ...,i_m\in \{1,2,...,r\}$, we have $$\sum_{j=1}^rO_{i_1j}...O_{i_mj} = \delta_{i_1...i_m},$$ where $\delta_{i_1...i_m} =1$ if and only if $i_1=...=i_m=1$ and $0$ otherwise.
Here are some of my thoughts. When $m=2$, $O$ can be any othogonal matrix, the desired equality is satisfied. When $m$ is an even number greater then two, we can very easy to prove that $O$ is those matrices with exactly one $1$ in every row and every column. Thus, $\alpha O$ is the matrix $\alpha$ after permuting its columns, which also leads to the desized equality. However, when $m\ge3$ is an odd number. I don't know whether $O$ can only be taken over theso permutation operater matrices. Alternatively, we can also try to prove or give a counter example of the question that whether every entry of $O$ must be non-negative. Also we can try to prove or give a counter example of the question that if there is a column or row of $O$ containing two non-zero entries is non allowed? Thank you for your attention.