Suppose tensor $U_{i\alpha\beta}$ with dimension $M*N*N$ satisfy following condition: $$U_{i\beta\alpha}=W^1_{\alpha\alpha'}W^2_{\beta\beta'}U_{i\alpha'\beta'}$$ where $W^1$ and $W^2$ are $N*N$ unitary matrix, which are independent of $i$.
Then, we define $\tilde{U}_{i\alpha\beta}\equiv V^1_{\alpha\alpha'}V^2_{\beta\beta'}U_{i\alpha'\beta'}$, where $V^1,V^2\in U(D)$, independent of $i$.
Is it possible to find $V^1,V^2$ s.t. $\tilde{U}_{i\alpha\beta}=\tilde{U}_{i\beta\alpha}$?
More generally, for $M*N*N*N\dots$ tensor $U_{i\alpha\beta\gamma\dots}$ satisfying following condition $$U_{iS(\alpha)S(\beta)S(\gamma)\dots}=W^1_{\alpha\alpha'}W^2_{\beta\beta'}W^3_{\gamma\gamma'}\dots U_{i\alpha'\beta'\gamma'\dots},$$ where $W^i\in U(D)$, can one find similar unitary transformation $V^i\in U(D)$, s.t. $$\tilde{U}_{i\alpha\beta\gamma\dots}=\tilde{U}_{iS(\alpha)S(\beta)S(\gamma)\dots}.$$ Here, $S$ is any permutation, and $\tilde{U}_{i\alpha\beta\gamma\dots}\equiv V^1_{\alpha\alpha'}V^2_{\beta\beta'}V^3_{\gamma\gamma'}\dots U_{i\alpha'\beta'\gamma'\dots}$