= Solution
In winding sector m write the lifted path as
$$
\phi(\tau)=\phi_0+\frac{2\pi m\tau}{\beta}+\eta(\tau),\qquad \eta(0)=\eta(\beta)=0.
$$
The cross term vanishes since $\int_0^\beta\dot\eta\,d\tau=0$. The action separates into $I(2\pi m)^2/(2\hbar^2\beta)$ and the same fixed-endpoint <Gaussian path integral> for every m. Neither term depends on the initial angle, so its integral supplies $2\pi$:
$$
\boxed{Z=2\pi Z_0\sum_m e^{-I(2\pi m)^2/(2\hbar^2\beta)},\qquad
Z_0=\int_{\eta(0)=\eta(\beta)=0}\mathcal D\eta\,
e^{-I\int_0^\beta\dot\eta^2d\tau/(2\hbar^2)}.}
$$
The <fixed-endpoint rotor fluctuation normalization> is
$$
\boxed{Z_0=\sqrt{\frac{I}{2\pi\hbar^2\beta}}.}
$$
It is the diagonal free-particle thermal kernel on the lifted line, or the amplitude density for a zero-displacement fluctuation bridge. It is not by itself the circle partition function or a separately countable angular zero mode. The integration over $\phi_0$ has already counted that mode. Quadraticity makes the saddle-plus-fluctuation evaluation exact.
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