The orbit category $\text{Orb}_G$ is defined in example 1.3.15 of Riehl.
It seems to me that the quotient group $G$ modulo $H$ corresponds to the full subcategory of $(\text{Orb}_G)^{\text{op}}$ on the object $G/H.$ Basically I'm wondering if this is correct, and if so, whether there is a way to reframe things to avoid the messy appearance of opposite categories. My reasoning is as follows:
The objects of $\text{Orb}_G$ are left $G$-sets $G/H$ (for subgroups $H \subseteq G$) and the arrows are the $G$-equivariant maps. Here $G/H = \{gH | g \in G \}$ has left action $g' \bullet gH = (g'g)H$ where $gH= \{gh | h \in H \}$ is a left coset.
It seems that the following conditions hold
For each arrow $G/H \overset{\alpha}\to G/K$ there exists $\gamma \in G$ such that $\alpha(gH)=g\gamma K$ for each $gH \in G/H.$ Here $\gamma^{-1} H \gamma \subseteq K.$
Given any $\gamma \in G$ with $\gamma^{-1} H \gamma \subseteq K$ we can define arrow $G/H \overset{\alpha^{\gamma}}\to G/K$ such that $\alpha^{\gamma}(gH)=g\gamma K$ for each $gH \in G/H.$
When $H$ is a normal subgroup we have $\gamma^{-1} H \gamma \subseteq H$ for each $\gamma \in G.$ In this case each arrow of $\text{Orb}_G$ with domain $G/H$ also has codomain $G/H,$ and the set of such arrows will be $\{\alpha^{\gamma} | \gamma \in G \}$ where $\alpha^{\gamma}(gH)=g\gamma H.$
The quotient group $G$ modulo $H$ has elements $gH$ and group multiplication defined so that $(gH) * (g'H)=\alpha^{g'}(gH)=(gg')H=\alpha^{g'} \alpha^{g} H.$
If we take the full subcategory $S$ of $\text{Orb}_G$ on the object $G/H$ its arrows will correspond to these endomorphisms $\alpha^{g},$ but these seem to compose in the opposite order to how multiplication works in $G$ modulo $H.$ Hence $G$ modulo $H$ (as a single object category) is equivalent to $S^{op}.$
Does this seem correct ?