Given is a topological space $X$ and a group $G \leq$ Aut($X$) with the property: for $\lambda \in G$ and $x \in X$, $\lambda(x)=x \Rightarrow \lambda=id_X$, also satisfying that the quotient map $q:X \to X/G$ has a continuous right inverse $s:X/G \to X$.
I want to show that if X is homogeneous (i.e. for all $x,y \in X$ there is $\alpha \in$ Aut($X$) s.t. $\alpha(x)=y$), then X/G is homogeneous as well.
I think I have found a proof for this when $G$ is a normal subgroup of Aut($X$), but it would be nice if it could be shown that this isn't a necessary condition.
Start with the quadrant $Q=\{(x,y): x\ge 0, y\ge 0\}\subset {\mathbb R}^2$. Let $Q_0$ denote the complement to $(0,0)$ in $Q$. Next, define the equivalence relation on $Q_0$ by $(x,0)\sim (0, x^{-1})$, $x>0$. Let $M$ denote the quotient space of $Q_0$ by this equivalence relation: The space $M$ is homeomorphic to the open Moebius band. Let $G={\mathbb R}$ with the addition as the group operation. The topological group $G$ acts continuously on $Q_0$: $$ t\star (x,y)= (e^tx, e^{-t}y). $$ This action descends to a continuous $G$-action on $M$. The action is clearly free. One verifies that the quotient space $W=M/G$ is homeomorphic to $[0,\infty)$. Moreover, take the ray $P=\{(1, y): y\ge 0\}\subset Q_0$ and let $R$ denote the projection of $P$ to $M$. Then the restriction of the $G$-quotient map $\pi: M\to W$ to the ray $R$ is a homeomorphism. (I leave verification of all these properties to you, proofs are straightforward.) The inverse $\pi^{-1}: W\to R\subset M$ also defines a right-inverse to $\pi: M\to W$. Lastly, the space $M$ is topologically homogeneous (since it is a connected manifold), but $W$ is not (since it is a manifold with nonempty boundary).
What goes "wrong" with this example is that the $G$-action on $M$ is free with Hausdorff quotient, but is not proper.