Explicite calculation of the map $m_g^* : H^0(G, \mathbb{Z}) \to H^0(G, \mathbb{Z})$ Let $G$ be a finite group and $g \in G$ be a non identity element. So we can write a map from $G$ to itself induces from $g$ i.e. $m_g : G \to G$  where $m_g(x) = g.x$. Again this $m_g$ induces a map in cohomology $$m_g^* : H^0(G, \mathbb{Z}) \to H^0(G, \mathbb{Z})$$
Question: $m_g^*(n_1, n_2, \cdots , n_{|G|}) = ?$
Note : Action of $G$ on $\mathbb{Z}$ is trivial.
Any help will be appreciated.
Thank you too much.
 A: It's true that if $X$ is a finite discrete space, then $H^0(X;\mathbb{Z}) \cong \bigoplus_{|X|} \mathbb{Z}$. However, it's easier for this exercise to back up a bit and see that $H^0(X;\mathbb{Z})$ is isomorphic to the group of maps of sets $X \to \mathbb{Z}$ (with pointwise addition for the group structure), i.e. $H^0(X;\mathbb{Z}) = \operatorname{Map}(X, \mathbb{Z})$. When $X = \{ x_1, \dots, x_n \}$ is finite, this is identified with $\mathbb{Z}^{\oplus n}$ by $\varphi \mapsto (\varphi(x_1), \dots, \varphi(x_n))$.
And more importantly, if $f : X \to Y$ is so map (which is automatically continuous), then the induced map $H^0(Y;\mathbb{Z}) \to H^0(X;\mathbb{Z})$ is given by pre-composition:
$$\varphi \in \operatorname{Map}(Y, \mathbb{Z}) \mapsto f^*\varphi := \varphi \circ f \in \operatorname{Map}(X, \mathbb{Z}).$$
So now for our finite group $G$, $H^0(G;\mathbb{Z}) \cong \operatorname{Map}(G, \mathbb{Z})$ (these are just maps of sets, not group homomorphisms). And if $\varphi : G \to \mathbb{Z}$, then $m_g(\varphi) = \varphi \circ m_g : x \mapsto \varphi(gx)$.
So given an enumeration $\{x_1, \dots, x_n\}$ of $G$ ($n = |G|$), you can view $H^0(X; \mathbb{Z}) \cong \mathbb{Z}^n$. If $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{Z}^n$, then $m_g^*\alpha$ is a permutation of $\alpha$, and you need to use the group law of $G$ to find out which one: $g x_i = x_{\sigma(i)}$ for some permutation $\sigma \in \mathfrak{S}_n$, and $m_g^*\alpha = (\alpha_{\sigma(1)}, \dots, \alpha_{\sigma(n)})$.
