Given a two groups $G, N$ with $N$ possibly abelian. Let $g,\alpha,\beta\in G$ and $x,y\in N$. I'm interested in studying functions $f:G\times G\times N\to G$ that satisfy this functional equation (where $g^\alpha=\alpha g\alpha^{-1}$ and using multiplicative notation).
$$f(g,\beta\alpha,xy)=f(g^\alpha,\beta,x)f(g,\alpha,y)$$
In general I would like to know if:
$q1)$ This identity is known, was long investigated, or has a name;
$q2)$ it is possible to derive from this identity some more concrete property of $f$ that can help to narrow the possible set of solutions of this functional equation;
$q3)$ Is it possible to describe/classify the set of solutions $f\in \mathcal F$?
Observation 1: I want $G$ to be non-abelian because otherwise we can just take a family of group homomorphisms $\{\phi_g\}_{g\in G}:G\times N\to G$ and set $F(g,\alpha, x):=\phi_g(\alpha,x)$. The set of solutions $f\in \mathcal F$ in the case $G$ is abelian is uninteresting an isomorphic to $ {\rm Hom}(G\times N, G)^G$
I did put a lot of effort${}^\dagger$ to understand the consequences of this equation but I really don't know what I'm missing.
In particular, if the previous is not feasible, I'd be happy to know the answer for the special case $N=\mathbb N$.
$q1^{nat}),\,q2^{nat}),\,q3^{nat})$ All the previous question but for the abelian case $N=\mathbb N$ ($G$ non-abelian).
My attempt${}^\dagger$:
The first trivial consequence is that $f(g,1_G,1_N)=1_G$. Define $f_g(\alpha,x):=f(g,\alpha,x)$. I had no luck deriving identities helpful enough to have a clue of what I'm doing: everything I got was the following
- If $g\in Z(G)$ then $f_{g}:G\times N\to G$ is a group homomorphism because for every $\alpha$ we have $g^\alpha=g$.
- For every $g$ the function $\phi_g(x)=f_g(1_G,x)$ is a group homomorphisms $\phi_g:N\to G$;
- In general we have $$f_g(\prod_{i=0}^j\alpha_i,\prod_{i=0}^jx_i)=\left(\prod_{i=1}^jf(g^{\displaystyle \prod_{k=0}^{i-1}\alpha_k},\alpha_i,x_i)\right) f_g(\alpha_0,x_0)$$
- From setting $\alpha=1_G$ I get $f_g(\beta,xy)=f_g(\beta,x)f_g(1_G,y)$ thus $$f_g(\beta,x_0\prod_{i=1}^j x_i)=f_g(\beta,x_0)\prod_{i=1}^j f_g(1_G,x_i)\quad (*)$$ and from that setting $x_0=1_N$ and $\beta=1_G$ one obtains $f_g(1_G,x^j)=f_g(1_g,x)^j$;
- Setting $\beta=1_G$ I get $f_g(\alpha,xy)=f_{g^\alpha}(1_G,x)f_g(\alpha,y)$ thus $$f_g(\alpha,(\prod_{i=1}^j y_i) y_0)=\prod_{i=1}^j f_{g^\alpha}(1_G,y_i)\cdot_G f_g(\alpha,y_0) \quad (**)$$
- Merging the two last eq.s $(*)\land (**)$ $$f_g(\beta,x^{j})=f_g(\beta,1_N)f_g(1_G,x)^j \\f_g(\alpha,x^{j})=f_{g^\alpha}(1_G,x)^jf_g(\alpha,1_N)$$ we obtain $$f_g(\alpha,1_N)f_g(1_G,x)^j=f_{g^\alpha}(1_G,x)^jf_g(\alpha,1_N)\quad(***)$$ thus $$\boxed{f_{g^\alpha}(1_G,x)=f_g(1_G,x)^{f_g(\alpha,1_N)}}$$
Context
I'm not sure if I should omit the context where this kind of equation came up because maybe it has nothing to do with the equation itself. The first equation came up, apparently, while enforcing the functoriality conditions of the component on arrows $F_1=f$ of an arbitrary functor $F\in[G^N\times N,G]$. At first I hoped I could parametrize the solution set of the f.eq. using the fact that $[G^N\times N,G]\simeq{\rm End}_{\bf cat}(G^N)$ as categories. I failed to compute one functor from the other establishing the bijection. Anyways establishing this probably wasn't going to answer $q1)$ and $q2)$.
Quick answer; these are actually semi-direct products; but weird ones.
And you are looking for homomorphisms,
$$ f : G \rtimes_{g^\alpha} N \to G\\ $$
Typically one has that $g^\alpha$ would be an isomorphism which depends on $N$ and not $G$. However, (if?) you can take a homomorphism from $N\to G$ this is equivalent.