Definition 1. Let $X$ be a set and $G$ be a group action on $X$. We define $[X/G]$ as a category with objects to be all elements $x\in X$ and $$\mathrm{Mor}(x,x')=\{g\in G:x'=g\cdot x\}.$$
Definition 2. For any set $S$, we define the category associated to $S$ to be $C_S$ with objects to be all elements in $S$ and morphisms to be identities. We call a groupoid $T$ (a category with morphisms are all isomorphisms) is equivalent to a set $S$ if there exists a equivalence $T\cong C_S$.
Question. Let $X$ be a set and $G$ be a group action on $X$. Then show that this is a free action iff $[X/G]$ is equivalent to a set.
If $[X/G]$ is equivalent to a set, then this is a free action. But I don't know the converse? May I pick $S=X/G$, the set quotient? But this seems not so right.
Thank you very much!
$[X/G]$ is equivalent to a discrete groupoid (the name for something like $C_S$) if and only if there is a set $S$ and a surjective map $F:X\to S$ such that for $x,x'\in X$, there exists an arrow $x\to x'$ if and only if $F(x)=F(x')$ in which case there exists a unique arrow $x\to x'$. That's because we can take $F$ to be the underlying function of the functor which provides equivalence $[X/G]\simeq C_S$ and apply the "fully faithful and essentially surjective" condition (in one direction) and in the other direction we just take this $F$ and easily lift it to a well-defined, fully faithful and (essentially) surjective functor.
The exists of such an $F$ automatically implies the action is free, since $gx=hx$ means $g,h$ are both arrows with the same domain and codomain in $[X/G]$ and there is a unique such arrow. Conversely if the action is free, the only thing we need to do is provide a surjective function $F:X\to S$ with the property that $F(x)=F(x')$ iff. there is an arrow $x\to x'$. Notice that the existence of an arrow $x\to x'$ is precisely the statement that $x\sim x'$ in the equivalence relation on $X$ induced by $G$. Therefore we want a surjective $F:X\to S$ with $x\sim x'$ iff. $F(x)=F(x')$. We see that $F$ is forced to be a quotient map and $S$ is forced to be (isomorphic to) the quotient set $X/G$.