= Solution
The <Pauli Z gate> exchanges the two <Hadamard basis> states: $Z|+\rangle=|-\rangle$ and $Z|-\rangle=|+\rangle$. Multiplying the displayed gate by the <Pauli X gate> gives
$$
U(\theta)X=|+\rangle\langle1|+e^{-i\theta}|-\rangle\langle0|.
$$
On the other hand,
$$
e^{-i\theta}ZU(-\theta)
=e^{-i\theta}Z\left(|+\rangle\langle0|+e^{i\theta}|-\rangle\langle1|\right)
=e^{-i\theta}|-\rangle\langle0|+|+\rangle\langle1|.
$$
The two operators agree, proving
$$
\boxed{U(\theta)X=e^{-i\theta}ZU(-\theta).}
$$
Also $U(\theta)=H\operatorname{diag}(1,e^{-i\theta})=J(-\theta)$ in the <J gate> convention. Keeping this angle sign explicit avoids confusing the two conventions for <equatorial qubit measurements>.
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