I've seen a couple of similar questions asking to verify computations of Bredon cohomology here and here, so I will ask one such question myself.

Let $\mathbb{Z}/2$ act on $S^3\subset \mathbb{C}^2$ by restriction of a permutation action on $\mathbb{C}^2.$ I wanted to compute Bredon cohomology $\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}}).$

I have a cell decomposition based on a decomposition of complex $1$-dimensional disk into $3$ cells: $\mathbb{D}=D\sqcup T\sqcup *.$ Here $T\sqcup *=S^1=\partial \mathbb{D}$ and $D$ is the interior of $\mathbb{D}.$ Then we have a decomposition of $S^3=\mathbb{D}\times S^1 \cup S^1\times \mathbb{D}$ into cells compatible with the $\mathbb{Z}/2$ action.

The fixed point set of an action is a circle given by $\{z_1=z_2\}\cap S^3\subset \mathbb{C}^2.$ Since the orbit category of $\mathbb{Z}/2$ consists of $*$ and $\mathbb{Z}/2$ there are the following equivariant chains: \begin{array}{|c|c|c|c|} \hline \operatorname{dim} &*& \mathbb{Z}/2 & \operatorname{cells corresponding to} \underline{C}_n(S^3)(\mathbb{Z}/2)\\ \hline 0 & \mathbb{Z} & \mathbb{Z} & * \times *\\ 1 & 0 & \mathbb{Z}\oplus\mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & T\times *,*\times T\\ 2 & 0 & \mathbb{Z}\oplus \mathbb{Z} \oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\\0\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 1\\0\end{pmatrix};\;\begin{pmatrix} 0 \\ 0\\1\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 0\\-1\end{pmatrix} & D\times *, *\times D, T\times T\\ 3 & 0 & \mathbb{Z}\oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & D\times T, T\times D\\ \hline \end{array}

So it seems that the cochains valued in $\underline{\mathbb{Z}}$ are:

\begin{array}{|c|c|} \hline \operatorname{dim} & \\ \hline 0 & \mathbb{Z}\\ 1 & \mathbb{Z}\\ 2 & \mathbb{Z}\\ 3 & \mathbb{Z}\\ \hline \end{array} Since $(T\times T)^*=0$ in cochains, we have $\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}.$ Differential $d_1$ is an isomorphism since $\partial(D\times *)=T\times *.$ It seems that $\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=H^*(S^3;\mathbb{Z}).$

It is a bit odd to me that the quotient is a homological sphere. Sure, the group $\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}$ since orientation is preserved, but maybe I've missed some $2$-torsion in lower degrees?