Can we say that $A$ is a complement for a group $G$?

Question

Let $A$ be a frobenius complement for a group $G$ i.e. $A$ act on $G$ by automorphism s.t. $C_A(g)=e$ for all nonidentity $g$.

Now, Action of $A$ can be linearly extended so that $A$ act on $F[G]$. As a result we can see $F[G]$ as a $F[A]$ module. Let $\chi$ be corresponding character of $A$ then

$$\ \chi(g) = \begin{cases} |G| & \textrm{ if $g=e$ } \\ 1 & \textrm{ if $g\neq e$ } \\ \end{cases} \ $$ as one can easily compute.

Now, I am asking the converse of this situation; if $A$ has a character $\chi$ s.t.

$$\ \chi(g) = \begin{cases} n & \textrm{ if $g=e$ } \\ 1 & \textrm{ if $g\neq e$ } \\ \end{cases} \ $$

then can we say that $A$ is a frobenius complement for a group of order $n$ ?

Note: $F$ can be taken as $\mathbb C$ complex field and I have asked it there but it think it is suitable for here.

Answer 1

As is so often the case, the answer is "not in general". Take $n=1+pq,$ where $p,q$ are primes with $q|p−1$. Let $A$ be a non-Abelian group of order $pq.$ Then $A$ is not a Frobenius complement, but $A$ does have a (complex) character which takes value $n$ on the identity and $1$ everywhere else. The character in question is the sum of the trivial character and the regular character.