= Solution
Put $D_\theta=\operatorname{diag}(1,e^{i\theta})$. Factoring the target operator gives
$$
U=D_\theta\otimes V,
$$
whereas $U'=I\otimes V$. Their difference is a <relative phase> on the first <qubit>, not one overall <global phase>.
For an explicit witness, choose the product input $|+\rangle\otimes|\chi\rangle$, with any normalized $|\chi\rangle$. The first-<qubit> outputs are respectively $(|0\rangle+e^{i\theta}|1\rangle)/\sqrt2$ and $|+\rangle$. A <Hadamard basis> measurement returns $+$ with <probabilities>
$$
\cos^2(\theta/2)\quad\text{and}\quad1.
$$
These differ because $\theta$ is not a multiple of $2\pi$. \b[The two <unitary operators> implement different <quantum channels>.] The <global phase> of the conditional target operation has become a physically observable phase on its control.
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