Let $G$ be a (discrete) group and $H\le G$ a subgroup of finite index. Then there is a transfer map 
$$tr\colon\thinspace H^\ast(H;M)\to H^\ast(G;M)
$$
in group cohomology, where $M$ is any $G$-module (see Brown's "Cohomology of groups", Chapter III).


I think this construction should be natural, in the following sense. Let $f\colon\thinspace G'\to G$ be a homomorphism such that $H':=f^{-1}(H)$ is of finite index in $G'$, and let $M$ be a $G$-module. Then the following diagram commutes, where the horizontal maps are transfers:
$$
\begin{array}{ccc}
H^\ast(H;M) & \to & H^\ast(G;M) \newline
\downarrow f^\ast & & \downarrow f^\ast \newline
H^\ast(H';f^\ast M) & \to & H^\ast(G';f^\ast M)
\end{array}
$$
Note that I do not want to assume that $(G':H')=(G:H)$ (however, I am willing to assume that $H\le G$ and $H'\le G'$ are normal, if necessary).



> 
> Does anyone know of a reference for this naturality?
> 
> 
> 



Naturality of the transfer in group cohomology
I don't believe this is true. Let $(G, H) = (\Sigma\_3, C\_3)$ and $f : C\_3 \to \Sigma\_3$. Then your square says that
$$H^1(C\_3;\mathbb{Z}/3) = \mathbb{Z}/3 \longrightarrow H^1(\Sigma\_3;\mathbb{Z}/3) = 0\longrightarrow H^1(C\_3;\mathbb{Z}/3) = \mathbb{Z}/3$$
is the identity, which is false.


