I was trying to answer by myself at this question. This is what I've done:
Let $e^\alpha_n$ be an $n$-cell of $X$ (I use the notations in the previous question). We have the attaching map $\chi^\alpha_n:\partial e^\alpha_n\cong S^{n-1}\rightarrow X_{n-1}$ where $X_{n-1}$ is the $(n-1)$-skeleton. So we have the maps
$\chi^{\alpha\beta}_n:S^{n-1}\rightarrow X_{n-1}\rightarrow\frac{X_{n-1}}{X_{n-1}-e^\beta_{n-1}}\cong S^{n-1}$. The differential on the cellular chain complex is defined by
$d(e^\alpha_n)=\sum_\beta\mathrm{deg}(\chi^{\alpha\beta}_{n})e^\beta_{n-1}$.
If $g\in G$ let's define $g\alpha$ as follows: $ge_n^\alpha=e^{g\alpha}_n$.
A Little computation show that the differential is a map of $G$-modules if and only if $\mathrm{deg}(\chi^{(g\alpha)(g\beta)}_n)=\mathrm{deg}(\chi^{\alpha\beta}_n)$.
Now we have the diagram \begin{matrix} H(S^{n-1}) &\rightarrow& H(S^{n-1})\\\\ \downarrow &&\downarrow\\\\ H(S^{n-1})&\rightarrow&H(S^{n-1}) \end{matrix} where the first orizzontal arrow is induced by $\chi^{\alpha\beta}_n$, the second orizzontal arrow by $\chi^{(g\alpha)(g\beta)}_n$ and the vertical by the multiplication by $g$. What I have to show is that the image of $1$ is the same in the two orizzontal arrow. So any of you could help me in showing this, please?
I asked this question also on mathoverflow