How to find out the number of conjugacy classes of complements in a special group extension? Let $U$ be an abelian group. Let $V:=\underbrace{U\times ... \times U}_{n\ \text{factors}}$.
Let $H:=C_n$ act on $V$ by cyclic permutation. Let this action be denoted by $\phi$.
Let $F:=V \rtimes_\phi C_n$.
Hence, $1\rightarrow V\stackrel{i}{\rightarrow} F\stackrel{p}{\rightarrow} H\rightarrow 1$ is a group extension.
Question:

Is there an easy way to see how many $H$-conjugacy classes of complements $V$ in $F$ are there?

Does one only have to find all group homomorphisms from $H$ to $F$ such that the sequence splits without computing the first cohomology group or does one have to compute the first cohomology group?
Thank you very much for the help.
 A: As I said in a comment, this follows from Shapiro's Lemma: if $H \le G$ are groups, and $U$ is an $H$-module (over ${\mathbb Z}$ as default), then $H^n(G,U^G) \cong H^n(H,N)$ for all $n \ge 1$. In your example, $H$ is trivial, so $H^n(G,U^G) = 0$ for all $n \ge 1$. This result extends naturally to the case when $U$ is an arbitrary non-abelian group.
It is not hard to give a direct proof of the result in your case, so here it is. There is no need to assume that $U$ is abelian.
Elements of $F = U \rtimes_\phi C_n$ have the form $vg^k$ with $v \in U^n$ and $0 \le k < n$ where, for $v = (u_1,u_2,\ldots,u_n)$, we have $gvg^{-1} = (u_2,u_3,\ldots,u_n,u_1)$.
Now a complement of $U^n$ in $F$ is generated by an element $vg$ with $(vg)^n = 1$. Since, $$(vg)^n = v (gvg^{-1}) (g^2vg^{-2}) \cdots (g^{n-1}vg^{-(n-1)}),$$
this is equivalent to $v=(u_1,\ldots,u_n)$ with $u_1u_2\cdots u_n=1$.
In that case, putting $c = (1,u_1,u_1u_2,\ldots,u_1u_2\cdots u_{n-1})$, we have $c^{-1}gc = c^{-1}(gcg^{-1})g = vg$, so the complements are conjugate.
