= Solution
Using the printed measurement-vector convention, the <unitary operator> is
$$
U(\theta)=\frac1{\sqrt2}\begin{pmatrix}1&e^{-i\theta}\\1&-e^{-i\theta}\end{pmatrix}
=H\operatorname{diag}(1,e^{-i\theta})=J(-\theta).
$$
Thus it is a <phase gate> followed by a <Hadamard gate>, with angle sign opposite to the parameter in the catalogued <J gate>. Its rows are the conjugate transposes of the orthonormal measurement vectors, so it is unitary. Right multiplication by the <Pauli Z gate> changes the sign of the second column, which is exactly the effect of left multiplication by the <Pauli X gate>, exchanging the two rows. Explicitly,
$$
U(\theta)Z=\frac1{\sqrt2}\begin{pmatrix}1&-e^{-i\theta}\\1&e^{-i\theta}\end{pmatrix}=XU(\theta).
$$
Similarly, right multiplication by the <Pauli X gate> exchanges columns, giving
$$
U(\theta)X=\frac1{\sqrt2}\begin{pmatrix}e^{-i\theta}&1\\-e^{-i\theta}&1\end{pmatrix}
=e^{-i\theta}Z\frac1{\sqrt2}\begin{pmatrix}1&e^{i\theta}\\1&-e^{i\theta}\end{pmatrix}.
$$
Therefore the exact identities, retaining the <global phase> in the second one, are
$$
\boxed{U(\theta)Z=XU(\theta),\qquad U(\theta)X=e^{-i\theta}ZU(-\theta).}
$$
These supply <Pauli-frame propagation along a measurement wire>.
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