= Solution
Assume $|\xi\rangle$ is normalized and the <Hilbert space> has dimension $d$. The rank-one <orthogonal projection> $P=|\xi\rangle\langle\xi|$ satisfies $P^\dagger=P$ and $P^2=P$.
For $d>1$, $P$ has <eigenvalue> zero on the orthogonal complement of $|\xi\rangle$, so it cannot be a <unitary operator>. The complementary <orthogonal projection> $I-P$ kills $|\xi\rangle$ and is not unitary in any positive dimension. In contrast, the <Householder reflection>
$$
R=I-2P
$$
is Hermitian and obeys $R^\dagger R=R^2=I-4P+4P^2=I$. Its <eigenvalues> are $-1$ along $|\xi\rangle$ and $+1$ on the orthogonal complement.
\b[For $d>1$, only $I-2|\xi\rangle\langle\xi|$ is unitary. In the exceptional one-dimensional case, $|\xi\rangle\langle\xi|=I$ is also unitary; its complement remains zero.]
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