Let $G$ be a finite and soluble group with a trivial center. We assume that $C_G(x_r)$ acts free of fixed points on $F(G)$ by conjugation and $C_G(x_r)F(G)$ is a Frobenius group with complement $C_G(x_r)$ and nucleus $F(G)$, where $x_r$ is an element of prime order. Suppose that $C_G(x_r)$ has even order. So $C_G(x_r)$ has a single $z$ element of order $2$ and satisfies: $$ f=zf^{-1}z\, \text{ for all } f\in F(G), \quad F(G) \text{ is abelian,} \quad \text{and } z\in Z(C_G(x_r)). $$
Consider another element $y \in G$ of odd order such that $C_{F(G)}(y)=1$.
I managed to ensure that there is a $\{2,|y|\}$-subgroup of $\operatorname{Hall}$ $H$ of $F(G)\langle z\rangle \langle y\rangle$ such that $y\in H$. I wonder if there is any result that guarantees that there is $f\in F(G)$ such that $[y,fzf^{-1}]=1$.