Definition. Let $G$ be a group and $X$ be any set. We may define a group action of $G$ on $X$ as map $\cdot: G\times X\to X$ such that $e\cdot x=x$ for all $x\in X$ and $g\cdot(h\cdot x)=gh\cdot x$ for all $g, h\in G$ and $x\in X$ (Here, of course, $g\cdot x$ is just a way of writing $\cdot(g, x)$).
Alternatively, we may equivalently say that a group action of $G$ on $X$ is a group homomorphism $G\to \text{Bijections}(X)$, where $\text{Bijections}(X)$ is the set of all the bijections of $X$ which makes a group under composition (This is the permutation group on $X$. I have avoided the use of the notation $S_X$).
Definition. Now suppose $G$ is any group and $X$ be a topological space. We may say that $G$ acts on $X$ if there is a map $\cdot:G\times X\to X$ such that $\cdot$ is an action when $X$ is considered as a set along with an extra condition that the map $x\mapsto g\cdot x:X\to X$ is continuous for all $g\in G$.
This may again be equivalently phrased as follows: A group action of $G$ on a topological space $X$ is a group homomorphism $G\to \text{Homeo}(X)$, where $\text{Homeo}(X)$ is the set of all the homoemorphism $X\to X$ under composition.
The "alternative descriptions" have a sense of uniformity. In fact, if $X$ is any object, then we may say that an action of $G$ on $X$ is a homomorphism $G\to \text{Aut}(X)$.
My Question is the following:
Suppose that $G$ is a topological group, that is, $G$ has a topology under which the binary operation and the inverse map are continuous. Let $X$ be a topological space. The standard definition of a continuous action of $G$ on $X$ is a continuous map $G\times X\to X$ which is also a group action when $X$ is thought of only as a set.
What I was wondering is can we give a topology on $\text{Homeo}(X)$ such that a continuous group homomorphisms correspond exactly to continuous actions of $G$ on $X$.