Given a finite group $G$, and a finite category $\mathcal{C}$, one can define the action of $G$ on $\mathcal{C}$ as a functor $A\_{\mathcal{C}}\colon G\to\mathbf{Cat}$, which takes the single object of $G$ (regarded as a category) to $\mathcal{C}$. Moreover, one can define the quotient $\mathcal{C}/G$ to be the colimit of $A\_{\mathcal{C}}$.
There is an explicit construction of the category $\mathcal{C}/G$: let's denote the orbit of an element $a$ by $Ga$. The object set of $\mathcal{C}/G$ is simply given by the orbits of the elements of $\mathcal{C}^{(0)}$.
To construct the set of morphism ${\mathcal{C}/G}^{(1)}$, one defines a relation $\leftrightarrow$ on $\mathcal{C}^{(1)}$ by saying $f\leftrightarrow g$ iff there is are decompositions $f = f\_1\circ\...\circ f\_n$ and $g=g\_1\circ\...\circ g\_n$, such that $G f\_i = G g\_i$ for all $i=1,...,n$.
This relation is clearly symmetric and reflexive. It is however, not transitive. So one defines $\sim$ to be the transitive closure of $\leftrightarrow$ and sets $\mathcal{C}/G^{(1)} := \mathcal{C}^{(1)}/\sim$.
My problem is that even though I can imagine why transitivity fails (i.e. given 3 morphisms $f,g,h$, one might find decompostions such that $f\sim g$ and $g\sim h$, but no decompositions such that $f\sim h$), I can't find an explicit example to demonstrate that case.


P.S.: As far as I know this might very well work too without the restrictions to finite groups and categories. But I haven't thought this through yet. So I restricted my question to the finite case.



Explicit construction of the quotient of a category by a group action
This question has been bugging me since it was posted, in part because I keep thinking you should have a homotopy action and should take the homotopy quotient. 


But anyway.... whew! I have an example which demonstrates why the relation is not transitive. The example consists of a category C with four objects $x,y, y'$, and $z$. The group in question is the cyclic group of order two $G = \mathbb{Z}/2 \mathbb{Z}$ and the action on C fixes the objects identically.


The category C is almost a free category. Here are the generating morphisms:


* morphisms $a\_i$ for $i = 0,1$ which go from $x$ to $y$,
* morphisms $a\_i'$ for $i = 0,1$ which go from $x$ to $y'$,
* morphisms $b\_i$ for $i = 0,1$ which go from $y$ to $z$, and
* morphisms $b\_i'$ for $i = 0,1$ which go from $y'$ to $z$.


The effect of $k \in G$ on $a\_i$, $a\_i'$, $b\_i$, or $b'\_i$ is to send $i$ to $i+k$ mod 2. There are two relations which we impose to define this category:


* $ b\_0 a\_0 = b\_0' a\_0'$, and
* $ b\_1 a\_1 = b\_1' a\_1'$.


Notice that this pair of identifications is compatible with the group action. This means there are exactly six morphisms from x to z. Now all of them are equivalent under the relation generated by $\leftrightarrow$, but this relation is not transitive. For example we have:


$$b\_1 a\_0 \leftrightarrow b\_1 a\_1 = b\_1' a\_1' \leftrightarrow b\_1' a\_0'$$


but there is no direct relation between $b\_1 a\_0$ and $b\_1' a\_0'$.


A similar construction works with G any group with the index $i \in G$ given by group elements. In this case the quotient category, as you defined it, is the `free walking commutative square'.


