Show that image of $res$ lies in $H^n(H,A)^{G/H}$ Let $G$ and $G^{\prime}$ be groups, $A$ and $A^{\prime}$ be $G$-module and $G^{\prime}$-module respectively, $C^n(G,A)$ be set of all maps from $G \times \cdots \times G$ ($n$ times) to $A$, $d_n :C^n(G,A) \rightarrow C^{n+1}(G,A)$ be coboundary operator, $H^n(G,A)$ be $n$ dimensional cohomology group. 
The group homomorphisms $\phi:G^{\prime} \rightarrow G$ and $\psi : A \rightarrow A^{\prime}$ are said to be compatible pair if $\psi$ is a $G^{\prime}$-module homomorphism when $A$ is made into $G^{\prime}$-module by means of $\phi$. This will induce a homomorphism from $C^n(G,A)$ to $C^n(G^{\prime},A^{\prime})$ by $f \mapsto \psi \circ f \circ \phi^n$. Since this homomorphism commutes with coboundary operator, this induces a homomorphism from $H^n(G,A)$ to $H^(G^{\prime},A^{\prime})$.
Suppose $H$ is a normal subgroup of $G$. For a fixed $g \in G$, let $\psi(a)=ga$ and $\phi(h)=g^{-1}hg$. ONe can check that $\phi$ and $\psi$ are compatible pair. One can check that the homomorphism $\theta_g$ from $H^n(H,A)$ to $H^(H,A)$ induced by $\phi$ and $\psi$ is automorphism and the quoteint group $G/H$ acts on $H^n(H,A)$ by $\theta_g$. 
Let $res : H^n(G,A) \rightarrow H^n(H,A)$ be the restriction homomorphism. I need to check that image of $res$ lies in $H^n(H,A)^{G/H}$, where $A^G=\{a \in A|ga=a ~\text{for all}~ g \in G \}$. For this, I need to check that $\theta_g(\bar{f|_H})=\bar{f|_H}$, where $\bar{f|_H}$ represents the class of $f|_H \in C^n(H,A)$. But I am unable to prove this because of following:
$\theta_g(\bar{f|_H})(h_1,\cdots,h_n)=\bar{\psi \circ f|_H \circ \phi^n}(h_1,\cdots,h_n)=\bar{gf|_H(g^{-1}h_1g,\cdots,g^{-1}h_ng)}=\bar{f|_H(h_1g,\cdots,h_ng)} $
Am I doing something wrong or missing something? 
 A: I did the calculation for $n=2$, and I am afraid I am going to have to leave it to you to try and generalize it! It is not so easy, because you have to find a coboundary. I have a feeling that there might be a proof using less calculation (maybe a dimension shifting argument?), but it is a while since I did much group cohomology!
Let $\phi \in Z^2(G,A)$. We have to prove that $\rho \in B^2(H,A)$, where
$\rho(h_1,h_2) = g\phi(g^{-1}h_1g,g^{-1}h_2g) - \phi(h_1,h_2)$.
Since $\phi$ is a $2$-cocycle, we have
$g\phi(g^{-1}h_1g,g^{-1}h_2g) = \phi(h_1g,g^{-1}h_2g)-\phi(g,g^{-1}h_1h_2g) + \phi(g,g^{-1}h_1g),$ so
$\rho(h_1,h_2) = \phi(h_1g,g^{-1}h_2g)-\phi(g,g^{-1}h_1h_2g) + \phi(g,g^{-1}h_1g)- \phi(h_1,h_2)$.
To show $\rho \in B^2(H,A)$, we need to find $\psi:H \to A$ such that 
$\rho(h_1,h_2) = h_1\psi(h_2) - \psi(h_1h_2)+\psi(h_1).$
You will find that $\psi(h) = \phi(g,g^{-1}hg)-\phi(h,g)$ works.
To check it, use the fact that $\phi$ is a $2$-cocycle to get
$h_1\psi(h_2) = \phi(h_1g,g^{-1}hg)-\phi(h_1,h_2g)+\phi(h_1,g) -\phi(h_1h_2,g) + \phi(h_1,h_2g)-\phi(h_1,h_2),$
and then the check is routine.
