Let $W$ be a quasi-primitive faithful and irreducible $H$-module with $H$ and $W$ of odd order. Suppose that the Fitting subgroup $F(H)$ is cyclic (so $H \le \Gamma(W)$ the semilinear group). Then each element of $W \setminus \{0\}$ generates a regular orbit under the action of $F(H)$.
I'm struggling with this, any idea? Of course $H$ must be solvable by Feit-Thompson Theorem. I don't figure out how to prove a subgroup of prime order in $F(H)$ that centralizes one element $x \in W\setminus \{0\}$ must centrlize every other non zero element, if this is the way to preceed.
Edit: quasi-primitive means that whenever $N$ is a normal subgroup of $H$, the restriction $W_N$ is homogeneous.
Here's my solution. Call $\mathbb{F}$ the base field of $V$. The restriction $V_{F(G)}$ is homogeneus and we can write $V_{F(G)}=fU$, so $U$ a faithful irreducible $F(H)$-module. But $F(H)$ is cyclic and so $\dim_{\mathbb{F}}(U)=1$, this means that $U \lesssim \mathbb{F}^{\times}$ and $F(G)$ acts as scalar multiplication on each irreducible constituent. So, if $v \in \mathbb{F}$ and $F(G)=<g>$ then $x^g=xc$ for a $c \in \mathbb{F}$. Being $V_{F(G)}$ homogeoneus $c$ is constant over all $F(G)$-irreducible constituent and then $F(G)$ acts as scalar multiplication on $V$ by a constant $c$ that has order $|F(G)|$ in $\mathbb{F}$. From this follows that every non zero element for a regular $F(G)$-orbit.
Once did it, I realize that this problem maybe isn't worth to be published to SE. However I cannot find a quicker way to prove it, as suggested by Derek Holt in a comment.