= Solution
Label the polygon vertices $v_0,\ldots,v_{2n-1}$ cyclically. Translation pairing of opposite sides identifies
$$
v_i\sim v_{i+n+1},\qquad v_{i+1}\sim v_{i+n},
$$
with indices modulo $2n$. The vertex classes are consequently the cosets of the subgroup generated by $n-1$ in $\mathbb Z/(2n)$, and their number is
$$
V=\gcd(2n,n-1)=
\begin{cases}1,&n\text{ even},\\2,&n\text{ odd}.\end{cases}
$$
The quotient is a <compact> <Hausdorff space>. An interior point has a disk neighbourhood, a paired-side point has two half-disks joined to a disk, and each vertex class has its incident sectors cyclically joined to a cone, topologically a disk. Thus it is a <connected> <closed surface> with an <orientation>. Its <Euler characteristic> is $V-n+1$, since there are $n$ paired edges and one face. Hence
$$
\boxed{g(X_n)=\frac{n+1-V}{2}=\left\lfloor\frac n2\right\rfloor.}
$$
The polygon interior and paired-side charts are <translation surface> charts with $\omega_n=dz$. Each corner angle is $(n-1)\pi/n$. For even $n$, all $2n$ corners meet, giving <cone angle> $2\pi(n-1)$; for odd $n$, each of the two classes contains $n$ corners, giving <cone angle> $(n-1)\pi$. Both are integral multiples of $2\pi$. A cone of angle $2\pi(k+1)$ has the local uniformizing coordinate $\zeta$ with $w=\zeta^{k+1}$; the <holomorphic one-form> is $(k+1)\zeta^k\,d\zeta$. Filling the vertices therefore supplies the <Riemann surface> structure and the <holomorphic one-form>, with
$$
\boxed{\begin{array}{ll}
n\text{ even}:&\text{one vertex class of zero order }n-2,\\
n\text{ odd}:&\text{two vertex classes, each of zero order }(n-3)/2.
\end{array}}
$$
An order-zero entry denotes a regular point, not an actual zero: for $n=2,3$ the surface is a <torus> and the form is nowhere zero. The <zero orders> otherwise sum to $2g-2$, as a check against the degree of the <canonical bundle>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-132-polygons.png]
{title=Opposite-side pairings: one vertex class in the octagon and two in the decagon}
For $n=4$, the form has one double zero and lies in the <stratum of holomorphic one-forms> $\mathcal H(2)$; for $n=5$, it has two simple zeros and lies in $\mathcal H(1,1)$. The <SL2R action on differentials> preserves zero multiplicities, as the local cone argument shows. Therefore
$$
\boxed{(X_4,\omega_4)\text{ and }(X_5,\omega_5)\text{ are not in the same }\mathrm{SL}_2(\mathbb R)\text{ orbit}.}
$$
Their equal <genus> and area do not distinguish the orbits; their different <strata of holomorphic one-forms> do.
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