= Solution
Retain the identification with the <left regular action>. A <group> <group automorphism> $\varphi$ of $G$ satisfies $\varphi L_g\varphi^{-1}=L_{\varphi(g)}$, so it normalizes $L(G)$ and fixes the identity point.
Conversely let a normalizing <permutation> $n$ fix the identity. Its conjugation on $L(G)$ induces a <group> <group automorphism> $\varphi$ characterized by $nL_gn^{-1}=L_{\varphi(g)}$. Evaluating at the identity gives $n(g)=\varphi(g)$, so the whole identity stabilizer in the <normalizer> is exactly $\operatorname{Aut}(G)$. For arbitrary $n$, put $t=n(1)$. Then $L_t^{-1}n$ fixes the identity and is an <group automorphism>; thus every element has a unique expression $L_t\varphi$.
The translation <subgroup> is normal in its <normalizer>, its intersection with the <group automorphism> <subgroup> is trivial, and the multiplication rule is
$$
(L_t\varphi)(L_u\psi)=L_{t\varphi(u)}\varphi\psi.
$$
Consequently
$$
\boxed{N_{\operatorname{Sym}(\Omega)}(G)\cong G\rtimes\operatorname{Aut}(G)=\operatorname{Hol}(G).}
$$
The action in this <semidirect product> is the natural evaluation of <group automorphisms> on $G$. This <normalizer> is the <holomorph>, realized as all <permutations> $x\mapsto t\varphi(x)$. The decomposition is established from the <regular group action> itself rather than only from an order comparison.
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