Recall that a CW-complex $X$ with an action of a group $G$ which permutes the cells (i.e., for any $g \in G$ and any cell $\sigma \subseteq X$, $g\sigma$ is a cell) is called a $G$-complex. If the action permutes the cells freely ($g\sigma = \sigma$ implies $g=1$), $X$ is a free G-complex.
Clearly, if $X$ is a free $G$-complex, then the $G$-action on $X$ is free (i.e., for any $g \in G$ and any $x \in X$, $gx = x$ implies $g=1$). A question that pops to my mind every once in a while is the following: is a $G$-complex with a free $G$-action a free $G$-complex? I see that if $g\sigma = \sigma$ for some nontrivial $g \in G$ and a cell $\sigma$, then $g$ has infinite order (for a finite group cannot act freely on a contractible space), but this doesn't seem to get me anywhere.