Can any group be identified with the fundamental group of a certain manifold?

Question

I'm a little confused. By covering theory, the fundamental group of a manifold is identified with the group of deck transformations of its universal covering. So if an automorphism has a fixed point, then it is the identity (by uniqueness of lifts). Which excludes the symmetric group (which has permutations different from the identity, which fix points).

Is my reasoning correct?

Answer 1

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).