= Solution
The <marked-state phase oracle> changes only the amplitude of $|x_0\rangle$. Since that amplitude in $|\psi\rangle$ is $1/2$,
$$
\boxed{I_{x_0}|\psi\rangle=|\psi\rangle-|x_0\rangle.}
$$
Also,
$$
\boxed{|\psi\rangle\langle\psi|x_0\rangle=\frac12|\psi\rangle.}
$$
It follows that $A|\psi\rangle=-|\psi\rangle$ and $A|x_0\rangle=|x_0\rangle-|\psi\rangle$, so
$$
A I_{x_0}|\psi\rangle
=A(|\psi\rangle-|x_0\rangle)
=-|x_0\rangle.
$$
The global minus sign has no observable effect. Prepare $|\psi\rangle=H^{\otimes2}|00\rangle$, query $I_{x_0}$ once, apply the fixed operator $A$, and perform a <quantum measurement in the computational basis>. The outcome is $x_0$ with probability one. This is <exact Grover search on four entries>.
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