A representation of a finite group where every nonzero vector has a trivial stabilizer What are the finite groups which admit a non-zero representation in char 0 where every non-zero vector has stabilizer equal to $\left<1\right>$? Cyclic groups of prime order is one obvious class.  Is there anything else?
 A: I think these groups have appeared on MO before.  A finite group $H$ has such a representation in characteristic zero if and only if $H$ occurs as a Frobenius complement. This is "well-known folklore" and appears in a book by D. Passman for example.
First, if $H$ is Frobenius complement in a Frobenius group $G$ with Frobenius kernel $K$,
any non-trivial minimal $H$-invariant subgroup $V$ of $K$ is an elementary Abelian $q$-group for some prime $q$ not dividing $|H|$, since $|H|$ and $|K|$ are comprime. Also $H$ acts faithfully on $V$. We can pass to an algebraic closure of ${\rm GF}(q)$ and lift the representation of $H$ to characteristic zero, and we obtain a complex representation of $H$ such that all non-identity elements act without the eigenvalue $1$.
Conversely, if $H$ has a complex representation with this last property, then we can reduce the associated module (mod $q$) for some prime $q$ not dividing $|H|$. Then we easily obtain a semidirect product $VH$ with $V$ an elementary Abelian normal $q$-subgroup such that $H$ is a Frobenius complement.
The structure of a (finite) Frobenius complement $H$  is reasonably well understood. For example, Burnside knew that if $p$ and $q$ are different prime divisors of $H$, every subgroup of $H$ of order $pq$ is cyclic. As mentioned in comments, all Sylow subgroups of $H$ are cyclic or (generalized) quaternion.
The only perfect Frobenius complement is ${\rm SL}(2,5).$ There is a Frobenius complement of order $63$ with center of order $3$. 
Later edit: This last group has a complex irreducible representation of degree $3$ in which no non-identity element has an eigenvalue $1$. In general, a Frobenius complement of odd order is metacyclic.
Even later edit: Here is a quick direct proof that any finite non-trivial group $G$ which has all subgroups normal necessarily has an irreducible complex representation (which may not be faithful) of the type in the question: take $\chi$ to be an irreducible non-trivial complex character of $G$. Let $N = {\rm ker} \chi$. Then by induction, we may suppose that $N = 1$, as $G/N$ has all its subgroups normal. Hence we may suppose that $\chi$ is faithful. For any $g \in G^{\#}
$, Clifford's theorem tells us that ${\rm Res}^{G}_{\langle g \rangle}(\chi)$ is a sum of $G$-conjugate irreducible characters of $\langle g \rangle$, since $\langle g \rangle \lhd G.$ None of these can be the trivial character (or they would all be trivial, since they are all $G$-conjugate, and that would put $g$ in ${\rm ker}\chi$). Hence $g$ acts without the eigenvalue $1$ in the representation affording $\chi$.
