= Solution
Write $F=F(G)$, $C=C_G(F)$ and $D=C\cap F=Z(F)$. Both $C$ and $D$ are normal in $G$. Suppose $C>D$. Choose a <minimal normal subgroup> $K/D$ of $G/D$ contained in $C/D$. In a finite <soluble group>, such a <minimal normal subgroup> is <elementary abelian>: the last nontrivial term of its <derived series> is an abelian <characteristic subgroup>, so minimality makes the whole <subgroup> abelian; its primary components and <subgroup> of elements killed by $p$ then force one prime and exponent $p$.
It follows that $K'\le D$. Moreover $K\le C$ centralizes $D\le F$, so $K'\le Z(K)$. Therefore $K$ is a <nilpotent group> of class at most two. Since $K$ is normal in $G$, the definition of the <Fitting subgroup> gives $K\le F$. But then $K\le C\cap F=D$, a contradiction. This proves the <Fitting subgroup is self-centralizing in soluble groups>:
$$
\boxed{C_G(F(G))\le F(G),\qquad C_G(F(G))=Z(F(G)).}
$$
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