Assume group $G$ acts on set $X$. My course notes say this:
Free action is a special case of faithful action. To have faithful action it suffices that any $g \neq e$ in $G$ moves something (but not necessarily everything) in $X$ to a different location, whereas free action requires that any $g \neq e$ has no fixed point.
My understanding: faithful actions cannot have all fixed points for a non-identity function because essentially each function must be unique, and the identity function has all fixed points, and therefore no other function can have all fixed points. In a free action, no function other than the identity function can have any fixed points.
Why can't there be two distinct functions, neither having any fixed points, which are equal on all points of $X$? Is that obviously impossible on its face?
My attempt to show why that's impossible: assume we have a free action of $G$ on $X$. Assume it's not faithful, so we have $g_1 \neq g_2 \in G$ such that for all $x \in X$, we have $g_1(x) = g_2(x)$.
If one of $g_1, g_2$ were the identity function, then we'd have $g_1(x) = g_2(x) = x$ for all $x \in X$, which by the definition of a free action would imply that both $g_1, g_2$ are the identity function, which violates our premise, so we can assume that neither is the identity function.
We have $g_2^{-1}(g_1(x)) = g_2^{-1}(g_2(x)) = g_2^{-1}g_2(x) = e(x) = x$, so $g_2^{-1} \neq e$ has a fixed point at $g_1(x)$, which contradicts the assumed free action. Therefore the free action must be faithful.
Does that make sense?