I feel a slight difficulty to conceive the following definition:
Setting: $G$ any group and $N \unlhd G$ a normal subgroup, $A$ is a $G$-module. The maps $c_g : N \rightarrow N$ and $m_g : A \rightarrow A$ are defined by $x \mapsto gxg^{-1}$ and $a \mapsto ag$ respectively.
The map $m_g$ is just a set map which inherit the $N$-morphism structure from: $(a \cdot (x)^{c_g})m_g = (a)m_g \cdot x$ ($a \in A, x \in N$). (usually referred as $(c_g, m_g)$ being a compatible pair). However, in that case the domain of $m_g$ is not the same as $A$ (the $N$-action is given by the twist by $g$). I am going to call it $A_g$. So $m_g : A_g \rightarrow A$ is an $N$-morphism.
Now $c_g$ induces $c^n_{g,N} : H^n(N,B) \rightarrow H^n(N,B)$ (for any $N$-module $B$ using contravariant functor $H^n(-,B)$) and $m_g$ induces $m^n_{g,N} : H^n(N,A_g) \rightarrow H^n(N,A)$ (using the covariant functor $H^n(N,-)$).
Now the action of $g \in G$ on $H^n(N,A)$ is given by $g^{(n)} := c^{n}_{g,A_g} m^{n}_{g,N}$.
${\mathrm{\bf Question.}}$ The $N$-action on $A$ that yield $A_g$ is sort of a local action. Even though $m^n_{g,N}$ is a group isomorphism (follows from functoriality and since $m_g$ is an $N$-isomorphism), I don't see why it should be natural. Isn't it necessary to have $m^n_{g,N}$ as a global equivariant functor to think this being an action on $H^n(N,A)$?