Are different versions of homogenous equivalent?

Consider these two definitions:

Definition 1. A topological space $X$ is called homogenous if for any $x,y\in X$ there is a homeomorphism $f:X\to X$ such that $f(x)=y$.

Definition 2. A topological space $X$ is called homogenous if there is a topological group $G$ acting continuously and transitively on $X$.

Of course definition 2 implies 1. The question is whether 1 implies 2? The natural choice is to look at $Homeo(X)$ with compact-open topology. However $Homeo(X)$ does not have to be a topological group. This holds when $X$ is compact Hausdorff or when $X$ is Hausdorff locally compact, locally connected. But in general there are examples of $X$ where the inverse operator on $Homeo(X)$ is not continuous.

However this does not imply that there is no other topological group acting continuously and transitively on $X$. Is it true? Or is there a counterexample?

Solution 1:

For 2 to 1 we have a topological group $G$ acting on our space $X$ ie. there is a continuous map $G \times X \rightarrow X$. Precomposing with the morphisms $(g,\operatorname{id}):* \times X \rightarrow G \times X$ gives us automorphisms of $X$ and transitivity of the action directly translates into the conditions on the automorphisms.

Conversely for 1 to 2 we note that the ordinary group $\operatorname{Aut}(X)$ always acts on $X$. But identifying ordinary groups with discrete topological groups this amounts to having a continuous map $$\begin{array}{rcl}\operatorname{Aut}(X)_\text{disc} \times X & \longrightarrow & X\\ (\phi, x) & \mapsto & \phi(x) \end{array}$$ This map is indeed continuous, since given an open subset $U \subseteq X$ its preimage takes the form $\bigcup\limits_{\phi\in\operatorname{Aut}(X)} \{\phi\}\times\phi^{-1}(U)$, which is open. Now again the conditions on the automorphisms translate into transitivity of the action.