 Solution

Assume $|\xi ⟩$ is normalized and the Hilbert space has dimension $d$. The rank-one orthogonal projection $P=|\xi ⟩⟨\xi |$ satisfies $P^{†}=P$ and $P^2=P$.

For $d>1$, $P$ has eigenvalue zero on the orthogonal complement of $|\xi ⟩$, so it cannot be a unitary operator. The complementary orthogonal projection $I-P$ kills $|\xi ⟩$ and is not unitary in any positive dimension. In contrast, the Householder reflectionis Hermitian and obeys $R^{†}R=R^2=I-4P+4P^2=I$. Its eigenvalues are $-1$ along $|\xi ⟩$ and $+1$ on the orthogonal complement.