= Solution
Choose $\alpha\in\Omega$. Since $G$ acts regularly, $g\mapsto g\alpha$ identifies $\Omega$ with the underlying set of $G$, and the given action becomes left multiplication $L_g(x)=gx$.
A <permutation> $c$ centralizing all $L_g$ satisfies
$$
c(x)=c(L_x(1))=L_x(c(1))=xc(1).
$$
Conversely every right translation commutes with every left translation. To make the multiplication convention explicit, define $R_g(x)=xg^{-1}$. Then $R_gR_h(x)=xh^{-1}g^{-1}=x(gh)^{-1}=R_{gh}(x)$, using rightmost-first composition. The map $g\mapsto R_g$ is injective and its image is the entire <centralizer>, since any possible $c(1)$ can be written $g^{-1}$. Thus
$$
\boxed{C_{\operatorname{Sym}(\Omega)}(G)\cong G.}
$$
This is <centralizer of a regular permutation subgroup>. If one instead writes right multiplication as $x\mapsto xg$, the resulting map is an antihomomorphism; including the inverse avoids suppressing that reversal. Choosing a different base point changes the identification but not the isomorphism type.
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