Why is $PT$ transformation not observable?

In the Preskill’s notes, QFT 1 page 70, he said:
Unlike charge conjugation $U_C$ and parity $U_P$, $U_{PT}$ is not an observable. For example
begin{align}
U_{PT},e^{itheta} psi=e^{-itheta},U_{PT}psi(x)
end{align}
so eigenvalues of $U_{PT}$ are not characteristic of (a well-defined property of) rays in Hilbert space.

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I don’t quit understand the statement.

(1) Did he mean $U_C$ and $U_P$ are observables? but these two are unitary operators rather than Hermitian.

(2) He said $U_{PT}$ are not characteristic of rays in Hilbert space. If he use the example of wave function,
$$U_{PT}psi(x)=psi^*(-x)$$
I would agree. But he used a phase factor which is changed under $U_{PT}$, I don’t understand since the phase factor is not a characteristic of rays in Hilbert space.