Let $M : V\otimes V \to V \otimes V$ be an $n^2 \times n^2 $ matrix ($V$ an $n$ dimensional vector space). Then define the operation
$$(M^D)^{ab}_{cd} = M^{db}_{ca}.$$
My hope is that this operation is basis independent, but I’m struggling to show it. A counter proof that it is basis dependent is also welcome.
Take $V$ to be $2$-dimensional, and $M$ to be the matrix which is all zeros except for $M_{12}^{12} = 1$. This matrix has trace $1$. However, the matrix $M^D$ has its only nonzero coefficient at $(M^D)_{11}^{22}$, and so $M^D$ has trace zero. So the operation is basis-dependent.