Cellular homology of the universal cover Let $X$ be a connected pointed CW complex. Let $\tilde{X}$ be its universal covering space and $G=\pi_{1}(X)$.
Lets denote $(C^{Cell}_{\ast}(\tilde{X}),d)$ the cellular chain complex associated to $\tilde{X}$.
By construction each $C^{Cell}_{n}(\tilde{X})$ is a free (left) $G$-module and the differential operators
$d:C^{Cell}_{n+1}(\tilde{X})\rightarrow C^{Cell}_{n}(\tilde{X})$ are (left) $G$-equivariant.
Question: Since each $C^{Cell}_{n}(\tilde{X})$ is a free (left) $G$-module, there is also a "natural" action of $G$ on $C^{Cell}_{n}(\tilde{X})$ on the right by multiplication. I was wondering if  $d:C^{Cell}_{n+1}(\tilde{X})\rightarrow C^{Cell}_{n}(\tilde{X})$ is also right $G$-equivariant ?
 A: As Benjamin Steinberg says, it does not work in general, and you can see the problem already for 1-dimensional complexes.  Suppose you take a Cayley graph for $G$.  This is a graph (or 1-dimensional CW-complex) with vertex set $G$ and edge set $G\times S$, where $S\subseteq G$ is a set of elements that generates $G$.  The edge $(g,s)$ joins the two vertices $g$ and $gs$.  I have set this up so that $G$ acts on the left: for $h\in G$, $h(g,s)=(hg,s)$ defines the action.  As we expected, the edge $h(g,s)$ joins the two vertices $hg$ and $hgs$.  Since the edge set and the vertex set are both free as $G$-sets, it is possible to make them into $G-G$-bisets.  But there isn't usually a way to do this that preserves the incidence relation between vertices and edges.  The first thing you might try is to define the right action on edges by $(g,s)k)=(gk,s)$.  But then the edge $(g,s)k$ joins the vertices $gk$ and $gks$, whereas the images under $k$ of the ends of the edge $(g,s)$ are $gk$ and $gsk$.  If the subset $S$ is closed under conjugation, then you could define $(g,s)k=(gk,k^{-1}sk)$, but if $S$ is not closed under conjugation (which it won't usually be) this won't work.  In the case of the universal cover of the wedge of two circles, $S$ is a set of two elements that freely generate the fundamental group.
In terms of $G$-sets, you can think about the vertices and directed edges as free $G$-sets, and the map taking a directed edge to its initial vertex as a $G$-map.  The left $G$-maps from $G$ to $G$ are exactly the right multiplications by elements of $G$, and unless $G$ is abelian, many of these will fail to commute with the right $G$-action.  So although you can make any free $G$-set into a $G-G$-biset, most left $G$-maps will not be $G-G$-bimaps.
