Let $V$ be a finite dimensional Hilbert space, then let $A: V\otimes V \to V\otimes V$ and we use index notation for the elements of the matrix $A$: $A^{\alpha \beta}_{\delta \gamma}= \langle e_{\alpha}\otimes e_{\beta}, A( e_{\delta} \otimes e_{\gamma})\rangle$ with $e_i$ an orthonormal basis of $V$. Define the following matrix operation $$(A^c)^{\alpha \beta}_{\delta \gamma} = A^{\delta \alpha}_{\gamma \beta}.$$ Does this operation preserve the usual matrix norm of $A$?
My idea for a proof is as follows:
We can formulate the operation $A^c = F(AF)^{Pt}$ where $F$ is the tensor flip on $V$ which can be easily seen to transpose the two top indices of $A$ when multiplied on the right and the two bottom indices of $A$ when multiplied on the right, and $A^{Pt}$ is the partial transpose of $A$ formally defined on pure tensors as $(M\otimes N)^{Pt} = M\otimes N^T$ and can be then seen to transpose the two rightmost indices of $A$. Since $F$ is clearly unitary, it doesn’t effect the matrix norm, and also for the partial transpose: $$\| (M\otimes N)^{Pt}\| = \|M\| \|N^T\| = \|M\|\|N\|= \|(M\otimes N)\|,$$ and hence also preserves the norm. Thus $\|A^c\|=\|A\|$.
The only parts I am unsure of is the partial transpose preserving the norm for maps on the tensor square that aren’t of the pure tensor form. Does this proof work?