= Solution
Write $C_U=|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes U$ for the <controlled unitary gate>. For a normalized <eigenvector> $|u\rangle$, the first <Hadamard gate> and <quantum phase kickback> give
$$
|0\rangle|u\rangle\longmapsto\frac{|0\rangle+|1\rangle}{\sqrt2}|u\rangle\longmapsto\frac{|0\rangle+e^{i\phi}|1\rangle}{\sqrt2}|u\rangle.
$$
The final <Hadamard gate> therefore produces the unentangled output
$$
|\chi_\phi\rangle|u\rangle,\qquad |\chi_\phi\rangle=\frac{1+e^{i\phi}}2|0\rangle+\frac{1-e^{i\phi}}2|1\rangle.
$$
The auxiliary system is unchanged, while its eigenphase becomes accessible through interference of the control <qubit>. The <special unitary group> assumption is not needed for this calculation; unitarity and the normalized <eigenvector> suffice.
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