My question refers to Tyler Lawson's answer to this question: Computing Bredon Cohomology of Z/2-spheres?

Namely, I have a problem with understanding 0'th degree. From my calculation it seems that $H^0(\mathbb{S}(2\sigma);M)=M(G)^G$. Here $G=\mathbb{Z}/2$ and $\sigma$ is a sign representation. Thus if we substitute $\mathbb{Z}$ with a sign action for $M(G)$ we will get that 0 degree cohomology is $0$ rather than $\mathbb{Z}$. Where is a mistake?