I've adapted the following proof from Wikipedia (here).
Let $\lambda: G→Sym(G)$ be the homomorphism induced by letting $G$ act on itself by left multiplication. Now let $H$ denote the normalizer of $\lambda(G)$ in $Sym(G)$.
We know that for each $f$ in $H$ and $g$ in $G$, there is an $h$ in $G$ such that $f\circλ(g) = λ(h)\circ f$.
The following is the sentence (verbatim) that confuses me.
If an element $f$ of $H$ fixes the identity of $G$, then for $1$ in $G$, $(f\circ λ(g))(1) = (λ(h)\circ f)(1)$, but the left hand side is $f(g)$, and the right side is $h$.
Basically, I just can't see why this is true. I'd appreciate it if anyone could spell this out. (I realize that I'm assuming this sentence is correct, but I think that it is because the rest of the article conforms to what I've seen in textbooks.)
$f \circ \lambda(g)$ is a bijection on the set $G$, which on $x \in G$ takes value $$ (f \circ \lambda(g)) (x) = f( g x). $$ Similarly $$ (\lambda(h) \circ f) (x) = \lambda(h) (f(x)) = h f(x). $$ So if $f(1) = 1$, and you take $x = 1$ in the above formulas, you get indeed $f(g) = h$. Taking now an arbitrary $x$ you have $$ f(g x) = f(g) f(x), $$ so that $f$ is an automorphism of $G$.