t^{a_1+\ldots + a_r+(n/n_1)c_1 + \ldots + (n/n_s)c_s}$$ giving $$\sum_{i=1}^r a_i + \sum_{j=1}^s \frac{n}{n_j}c_j \equiv 0 \pmod{n}.$$ The fact that the data set ${\operatorname{{\mathcal D}}}$ has genus $g$ follows easily from the multiplicativity of the orbifold Euler characteristic for the orbifold covering $S_g \to {\operatorname{{\mathcal O}}}$: $$\frac{2-2g}{n} = 2-2g_0 + r\left(\frac{1}{n}-1 \right) + \sum_{j=1}^s \left(\frac{1}{n_j} - 1 \right).$$ Thus, $h$ gives a $(n,r)$-data set $${\operatorname{{\mathcal D}}}= (n,g_0, (a_1,a_2,\ldots, a_r);(c_1,n_1),(c_2,n_2),\ldots, (c_s,n_s))$$ of genus $g$, and hence $({\operatorname{{\mathcal D}}}, {\operatorname{{\mathbb O}}}_t)$ forms a permuting $(n,r,k)$-data set.

Consider another permuting $(n,r,k)$-action $t'$ in the equivalence class of $t$ with a distinguished fixed point set ${\operatorname{{\mathbb P}}}(t') = \{P_1',P_2',\ldots,P_r'\}$. Then by definition there exists an orientation-preserving homeomorphism $\phi \in {\operatorname{Mod}}(S_g)$ such that $\phi(P_j) = P_j'$ for all $j$ and $\phi t\phi^{-1}$ is isotopic to $t'$ relative to ${\operatorname{{\mathbb P}}}(t')$. Therefore, $\theta_{P_j}(t) = \theta_{P_j'}(t')$, for $1\leq j\leq r$, and since ${\operatorname{{\mathbb O}}}_t = {\operatorname{{\mathbb O}}}_{t'}$, the two actions will produce the same permuting $(n,r,k)$-data sets.

Conversely, given a permuting $(n,r,k)$-data set $({\operatorname{{\mathcal D}}}, {\operatorname{{\mathbb O}}}_{{\operatorname{{\mathcal D}}}})$, we construct the orbifold ${\operatorname{{\mathcal O}}}$ and a representation $\rho : \pi_1^{orb}({\operatorname{{\mathcal O}}}) \to C_n$. Any finite subgroup of $\pi_1^{orb}({\operatorname{{\mathcal O}}})$ is conjugate to one of the cyclic subgroups generated 