= Solution
The upper-left block of $U$ is the planar <rotation matrix> $R(\pi/2)$, and $U$ fixes the third coordinate. Therefore
$$
U^p=\begin{pmatrix}\cos(p\pi/2)&-\sin(p\pi/2)&0\\\sin(p\pi/2)&\cos(p\pi/2)&0\\0&0&1\end{pmatrix},\qquad U^4=I.
$$
The diagonal entries of $V$ square to one, so $V^2=I$. Direct conjugation gives $VUV=U^{-1}$, hence $VU^pV=U^{-p}$ for every integer $p$. Consequently
$$
\boxed{U^pV=VU^{-p}=VU^{4-(p\bmod4)}.}
$$
When $p\bmod4=0$, the last exponent is four, which is equivalent to zero because $U^4=I$.
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