Consider a $4 \times 4$ Haar-random unitary $U$. By the right translational invariance of the Haar measure, for any $4 \times 4$ unitary $A$, $U A$ is still a Haar-random unitary.
Now, consider two $4 \times 4$ Haar-random unitaries, $U$ and $V$, and consider the following $16 \times 16$ matrix $Q$:
\begin{equation} Q = (U \otimes V) ~(I_{2 \times 2} \otimes A \otimes I_{2 \times 2}), \end{equation}
where $A$ is a $4 \times 4$ unitary and $I_{2 \times 2}$ is the $2 \times 2$ identity matrix.
Can it be written as a tensor product of two $4 \times 4$ Haar random unitaries? If not, is $Q$ a $16 \times 16$ Haar random unitary?
Note that since we cannot directly multiply $A$ with either $U$ or $V$, the translational invariance of the Haar measure cannot be directly used.