let $X$ be a topological space on which a group $G$ acts.
1) is it true that this action always induces an homomorphism $G\rightarrow Aut(X)$?
My guess is no. because i think the induced homomorphism would be the map that associates to $g\in G$ the map $\phi_g:X\rightarrow X; \, x\mapsto gx$. Here $\phi_g$ is clearly injective but it need not be surjective unless the action is transitive. Hence $\phi_g\not \in Aut(X)$.
2)Is it true that if $G$ is finite then the action is always transitive?
Best Answer
Following Mariano's suggestion, I'm turning my comments into an answer.
Recall that a left action of $G$ on $X$ can be described by a map $\alpha: G \times X \to X$ satisfying
Combining 1. and 2. gives in particular $\alpha(g^{-1},\alpha(g,x)) = \alpha(e,x) = x = \alpha(g^{-1},\alpha(g,x))$. Writing $\phi_{g} = \alpha(g,\cdot)$ we see that $\phi_{g^{-1}} \circ \phi_{g} = \operatorname{id}_{X} = \phi_{g} \circ \phi_{g^{-1}}$, so $\phi_{g}: X \to X$ is a bijection of $X$ with inverse $\phi_{g^{-1}}$, no matter if the action is transitive or not. Moreover, $\phi_{gh} = \phi_{g}\circ\phi_{h}$ follows immediately from 2. and means that $\phi: G \to \operatorname{Aut}{(X)}$ is a homomorphism.
On the other hand, given a homomorphism $\phi: G \to \operatorname{Aut}{(X)}$ we get an action $\alpha$ by setting $\alpha(g,x) = \phi_{g}(x)$. It is not difficult to check that the correspondence $\alpha \leftrightarrow \phi$ gives mutually inverse bijections between the set of actions $G \times X \to X$ and the set of homomorphism $G \to \operatorname{Aut}(X)$.
As for the second question, there is no reason for a finite group to act transitively on a set. It can't act transitively if $|G| \lt |X|$, so for instance letting $X = G \cup \{pt\}$ where $G$ acts on itself by left translations and fixes the point $pt$, we have an action that isn't transitive.