Here is how it goes.
Let $B$, be a space nice enough to have a (simply connected) universal cover, say $B$ is connected, locally connected and semi-locally simply connected. Let $(X,x_0)\to (B,b_0)$ be its universal cover.
Take a loop $\gamma: (S^1,1)\to (B,b_0)$ then you can lift $\gamma$ to a path $\overline{\gamma}: I\to X$ that projects to $\gamma$. Now $\overline{\gamma}(1)$ is an element of $X_{b_0}$. You can use then the following theorem.
Let $(Y,y_0)\to (B,b_0)$ be a (path) onnected and locally path connected space over $B$ and $(X,x_0)\to (B,b_0)$ is a cover of $B$, then a lift of $(Y,y_0)\to (B,b_0)$ to $(Y,y_0)\to (X,x_0)$ exists iff the image of $\pi_1(Y,y_0)$ inside $\pi_1(B,b_0)$ is contained in the image of $\pi_1(X,x_0)$ inside $\pi_1(B,b_0)$
Use the previous theorem with $(Y,y_0)=(X,\overline{\gamma}(1) )$.
This tells you that there exists a covering map $X\to X$ sending $x_0$ to $\gamma(1)$.
It is easy to see that this map depends only on the homotopy class of $\gamma$ using the following result
Let $(X,x_0)$ be a cover of $(B,b_0)$ and $Y$ be a connected space over $B$. If two liftings of $Y\to B$ to $Y\to X$ coincide at some $y_0$ in $Y$, the they're equal.
This tells you that if $\overline{\gamma}(1)=\overline{\tau}(1)$ then the two morphisms $X\to X$ you get, coincide.
Moreover, using the inverse of $\gamma$, you see that the morphisms $X\to X$ you get are automorphisms.
This gives you a well defined map $\pi_1(B,b_0)\to \text{Aut}_B(X)$.
Using what I said before, it is easy to see that it is an isomorphism.
Yes it is. Let $X$ be a topological space with universal cover $\widetilde{X}$ and covering map $p : \widetilde{X} \to X$. A homeomorphism $f : \widetilde{X} \to \widetilde{X}$ is called a deck transformation of $p$ if $p\circ f = p$; that is, $f$ preserves the fibres of $p$ so if $y \in p^{-1}(x)$, $f(y) \in p^{-1}(x)$. The set of all deck transformations of $p$ is denoted $\operatorname{Deck}(p)$ and forms a group under composition. The quotient of $\widetilde{X}$ by $\operatorname{Deck}(p)$ is $X$. Moreover, $\operatorname{Deck}(p) \cong \pi_1(X)$.
A good reference for this material is Hatcher's Algebraic Topology.
Best Answer
If $M$ is a manifold and $\hat M$ its universal cover, $\hat M$ is endowed with a differentiable structure such that $p:\hat M\rightarrow M$ is differentiable. For every $x\in M$, there exists a neighborhood $U$ of $x$ such that $U$ is a domain chart $\phi:U\rightarrow\mathbb{R}^n$ and $p^{-1}(U)=\bigcup V_i$ and $p_{\mid V_i}\rightarrow U$ is an homeomorphism, the differentiable structure is defined by supposing that $p_i$ is a diffeomorphism which induces the chart $\phi\circ p_i$ on $\hat V_i$.
Now look at the action of an element $\gamma\in\pi_1(M)$ on $\hat M$, i.e. a covering transformation. We will prove that this action is already smooth. Let $p(y)=x$. There exist $i$ and $j$ such that $y\in V_i$ and $\gamma . y\in V_j$, However, the map $(\phi\circ p_j) \circ \gamma \circ (\phi\circ p_i)^{-1}$ is the identity, this implies that action of $\gamma$ is differentiable.
This proves that every covering transformation is smooth.