In this question here, I asked if there could exist a $U \in U(4)$ such that $U$ itself was not the tensor product of two matrices, but such that $U(A \otimes B)U^{-1} = A' \otimes B'$ for all $A,B \in U(2)$. It was shown that the following matrix was such an example:
$F = \begin{bmatrix} 1&0&0&0\\ 0&0&1&0\\0&1&0&0\\0&0&0&1 \end{bmatrix}$.
This was nice because it satisfied the following nice condition for me: left multiplication by $F$ would take a simple tensor to a non-simple tensor, but conjugation by $F$ preserved simple tensors.
Now I'm wondering if something else might be true as well (or if there is some obvious reason that it can't be): does there exist a matrix $U \in U(4)$ such that again,
(i) $U(A \otimes B)U^{-1} = A' \otimes B'$ for all $A,B \in U(2)$, but also
(ii) $U(u \otimes v) \neq u' \otimes v'$ for some $u \otimes v \in \mathbb{C}^2 \otimes \mathbb{C}^2$ and for any $u' \otimes v' \in \mathbb{C}^2 \otimes \mathbb{C}^2$.
Namely is there an immediate reason why there cannot exist such a matrix $U$ which preserves simple tensors of operators under conjugation, but maps some simple tensor in the vector space to a non-simple tensor?