of a group $G$ on a set $X$ is free if for every $x \in X$ and every $g\in G$, the equation $g x = x$ implies $g=1_G$. Equivalently, an action is free when for any pair of elements $x,y \in X$, there is at most one group element $g \in G$ such that $g x = y$.

Any group $G$ acts freely on itself by multiplication $\cdot : G \times G \to G$, which is called the (left) regular representation of $G$.

For any set $X$ equipped with a transitive action$* : G \times X \to X$, the group $Aut_G(X)$ of $G$-equivariant automorphisms of $X$ (i.e., bijections$\phi : X \to X$ commuting with the action of $G$) acts freely on $X$. In particular, suppose $\phi \in Aut_G(X)$ is such that $\phi(x) = x$ for some $x\in X$, and let $y\in X$ be arbitrary. By the assumption that $G$ acts transitively, there is a $g \in G$ such that $y = g*x$. But then $G$-equivariance implies that $\phi(y) = \phi(g*x) = g*\phi(x) = g*x = y$. Since this holds for all $y\in Y$, $\phi$ must be equal to the identity $\phi = id_X$, and therefore $Aut_G(X)$ acts freely on $X$.