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.
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.
Best Answer
Let $M$ be a manifold with $\pi_1(M)\cong S_n$
You're conflating the action of the fundamental group on the universal cover of $M$ with the action of $S_n$ on an $n$-element set. It's possible for one of these actions to have a fix point, while the other does not.
It is indeed true that a non-identity element has no fixed point. This doesn't tell us anything about the group though: for any group $G$, if we let $G$ act on itself via left multiplication, every non-identity element has no fixed point. And this action is actually (equivariantly isomorphic to) the action of the fundamental group on each fiber of the universal cover (and deck transformations preserve the fibers).