Yet another question "I compute Bredon cohomology of something and I am not sure, whether it is correct".

So I am taking a sign representation $\sigma$ of cyclic group of order 4, $C_4$. Then I compactify $\sigma$ to get $\mathbb{S}^\sigma$ with two fixed 0-cells and one 1-cell of type $C_2$. Then I am taking $\underline{\mathbb{Z}}$, constant $\mathbb{Z}$-coefficients. It seems that $$ C_{C_4}^0(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=\mathbb{Z}\oplus\mathbb{Z} \\ C_{C_4}^1(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=0, $$ therefore $H_{C_4}^0(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=\mathbb{Z}\oplus\mathbb{Z}$ and $H_{C_4}^1(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=0$.

Degree 1 looks ok for me, but I am worried about degree 0. Here I calculated similar thing for $C_2$: Bredon cohomology of $\mathbb{S}^\sigma$ and degree 0 answer is different. But it should not change, since sign action of $C_4$ factors through sign action of $C_2$.

Also, since I am using constant coefficient system, $H_{C_4}^*(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=H^*(\mathbb{S}^{\sigma}/C_4;\mathbb{Z})$ (unless it is true only for cyclic groups of prime order) - so degree 0 cohomology should be only one $\mathbb{Z}$, as I am taking non-equivariant cohomology of a contractible space.

So is my answer correct?