For the circle $S^1$, we shall consider the ${\Bbb Z}/3{\Bbb Z}$-torsors. I know the isomorphism class of ${\Bbb Z}/3{\Bbb Z}$-torsors are classified by $H^1(S^1, {\Bbb Z}/3{\Bbb Z}) = {\mathrm{Hom}}(\pi_1(S^1), {\Bbb Z}/3{\Bbb Z}) = {\mathrm{Hom}}({\Bbb Z}, {\Bbb Z}/3{\Bbb Z}) = {\Bbb Z}/3{\Bbb Z}$.
I would like to realise these three torsors as visible manifolds which are non-isomorphic to each other. Apparently, $0 \in {\Bbb Z}/3{\Bbb Z}$ corresponds to the three disjoint union of $S^1$'s. I try to realise the ${\Bbb Z}/3{\Bbb Z}$-torsor of $S^1$ corresponding to $1 \in {\Bbb Z}/3{\Bbb Z}$.
My first attempt is to put some exotic structure to the connected Galois covering $Y \to S^1$, where ${\mathrm{Gal}}(Y/S^1) \cong {\Bbb Z}/3{\Bbb Z}$. Because $\pi(S^1) \cong {\Bbb Z}$, $Y$ is uniquely determined. Locally over $S^1$, $Y$ has three connected components which are disjoint from each other. I wish if $Y$ itself would be a ${\Bbb Z}/3{\Bbb Z}$-torsor. However, I know this is wrong, for there is no information of the character $\chi \colon \pi_1(S^1) \to {\Bbb Z}/3{\Bbb Z}$ added in the construction $Y$ which is a mere Galois covering. That is, $Y$ is actually a manifold, but cannot be a ${\Bbb Z}/3{\Bbb Z}$-torsor.
Q. What are connected manifolds other than $Y$ realised as ${\Bbb Z}/3{\Bbb Z}$-torsors of $S^1$?
$\newcommand{\Ints}{\mathbf{Z}}$The diagram shows the two non-trivial $\Ints/3\Ints$ torsors. The total space of each is a circle, divided into thirds. Each third is color-coded consistently, the base circle is in the middle, and the projection mappings are the obvious "squeezings" of the fibres to points, which send each arc onto the base circle.
The mapping carrying one torsor to the other in your linked question may be viewed as a suitable involution of the total space, e.g., complex conjugation of the unit circle in the complex plane, a.k.a. the inverse mapping $z = e^{i\theta} \mapsto \overline{z} = e^{-i\theta} = \frac{1}{z}$.