Q:

Does $\pi_1(X, x_0)$ act on $\tilde{X}$?

Let $X$ be a path connected, locally path connected space and let $p:\tilde{X} \to X$ be a covering map. Let $x_0\in X$. Then we have a natural right action of $\pi_1(X, x_0)$ on the fibre $p^{-1}(x_0)$ given by $x'_0\cdot [\gamma] =\tilde{\gamma} (1)$. Where $\tilde {\gamma} $ is the unique lift of $\gamma$ beginning at $x'_0$.
However, is there an action of this group on all of $\tilde {X}$? 

A:

As you noted in your question, $\pi_1(X,x_0)$ acts on $p^{-1}(x_0)$ by permuting the points (i.e. the lifts of $x_0$) around. But if you're looking for a group which acts on the entire space $\tilde X$ (which I believe is your question?) then you'd want the group of deck transformations $G(\tilde X)$. This group is the collection of all covering space isomorphisms $f:(\tilde X, \tilde x_0)\to (\tilde X,\tilde x_0)$, i.e. homeomorphisms of $\tilde X$ which satisfy $p=p\circ f$. (So intuitively, a homeomorphism $f$ is an element of $G(\tilde X)$ if the projection or "shadow" of $\tilde X$ onto $X$ looks just like the projection/"shadow" of the 'homeomorphed' space $f(\tilde X)$ onto $X$). 
In general, $\pi_1(X)$ and $G(\tilde X)$ are not isomorphic. But if $\tilde X$ is the universal cover then they are. (See, for instance, Proposition 1.39 of Hatcher to see a more general relationship between $G$ and $\pi_1$.) So in that case, the action of $\pi_1$ on the elements of $p^{-1}(x_0)$ does indeed coincide with an action on all of $\tilde X$.

