Embedding of $G$ naturally in symmetric group and its normalizer

Question

Let $G$ be a finite group of order $m$, and for each $g\in G$, let $L_g:G\rightarrow G$ be the permutation $x\mapsto gx$. Then $\{L_g:g\in G\}$ is a subgroup of $S_m$ isomorphic to $G$.

Consider any $\sigma\in S_m$. Then for any $g\in G$, $$(\sigma L_g\sigma^{-1})(\sigma(x))=\sigma(gx).$$ If $\sigma:G\rightarrow G$ is automorphism, then RHS in above equation is $\sigma(g)\sigma(x)$, which is $L_{\sigma(g)}(\sigma(x))$. So $$\sigma L_g\sigma^{-1}=L_{\sigma(g)}.$$ This implies that $\sigma$ normalizes $\{ L_g:g\in G\}$.

I want to show that if $\sigma\in S_m$ normalizes $\{ L_g:g\in G\}$ then $\sigma$ must be an automorphism of $G$. I was unable to see this through above computations. Any hint for this?

Answer 1

Of course it is not true, as if $x \ne 1$, then $L_{x}$ normalises $\{ L_g:g\in G\}$, as it lies in it, but it is clearly no automorphism of $G$, as it does not fix $1$.

Perhaps the confusion stems from the fact it would be more appropriate to take, instead of $S_{m}$, the group $S_{G}$ of permutations of the set $G$.

Instead, note that the normaliser $N$ of $\{ L_g:g\in G\}$ in $S_{G}$ factorises as $$ T \cdot \{ L_g:g\in G\}, $$ where $T$ is the stabiliser of $1$ in $N$ (this is because $\{ L_g:g\in G\}$ is a transitive subgroup of $N$), and then prove that $T = \operatorname{Aut}(G)$.

Spoiler

Suppose $\sigma \in N$, and $\sigma(1) = 1$. Then for each $g \in G$ there is $x \in G$ such that $$\sigma L_{g} \sigma^{-1} = L_{x}.\tag{eq}$$ Apply both sides of (eq) to $1$ to see that $\sigma(g) = x$, and then apply again both sides of (eq) to $\sigma(h)$, for $h \in G$, to see that $$\sigma(g h) = \sigma(g) \sigma(h).$$