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: $$ f=zf^{-1}z\ \text{for all } f\in F(G),\ \ F(G)\text{ is abelian and } z\in Z(C_G(x_r)). $$
I want to show that $G=C_G(z)F(G)$.
I was able to show that $\langle z\rangle F(G)\ {\rm char}\ C_G(x_r)F(G)$ and in this way I want to show that $C_G(x_r)F(G)\trianglelefteq G$ to use Frattini's argument and conclude the proof, but I still haven't been able to show that $C_G(x_r)F(G)\trianglelefteq G$.
Let $F = F(G)$ and let $g \in G$. We show first that $g^{-1}zgz \in C_G(F)$. Let $f \in F$. Then, since $gf^{-1}g^{-1} \in F$, we have $zgf^{-1}g^{-1} =gfg^{-1}z$, and so: $$g^{-1}zgzf = g^{-1}zgf^{-1}z = g^{-1}z(gf^{-1}g^{-1})gz = g^{-1}gfg^{-1}zgz = fg^{-1}zgz,$$ and $g^{-1}zgz \in C_G(F)$ as claimed.
But, since $G$ is solvable, $C_G(F) \le F$, so $g^{-1}zgz \in F$, and hence $g^{-1}zg \in \langle z \rangle F$.
Now $\langle z \rangle \in {\rm Syl}_2(\langle z \rangle F)$, and so $z$ and $g^{-1}zg$ are conjugate by an element $f \in F$, i.e. $g^{-1}zg = f^{-1}zf$, so $gf^{-1} \in C_G(z)$ and hence $g \in C_G(z)F$.