If G is a group and A is a subset of G, the normaliser of A in G can be defined as either
(1) $N_G(A) = \{g \in G\ |\ gag^{-1} \in A, \forall a \in A\}$
(2) $N_G(A) = \{g \in G\ |\ gAg^{-1} = A \}$ where $gAg^{-1} = \{ gag^{-1}\ |\ \forall a \in A\}$
I want to prove definition (1) $\implies$ definition (2) i.e. that the sets defined in (1) and (2) are the same
If A is finite then the fact that $f_g: A \to A,\ a \mapsto gag^{-1}\ (g \in N_G(A))$ is an injective function implies that $f_g$ is bijective (since an injective map from a finite set to itself is bijective).
However an injective map from an infinite set to itself isn’t necessarily bijective. So my question how would one go about proving this in the case that A is infinite?
(1) doesn't imply (2) (where by (1) and (2) I think you mean the condition such that an element of $G$ is in the subset)
Consider the following counterexample: $G=\Bbb{Q}\rtimes \Bbb{Q}^\times$ As a set we have $G=\Bbb{Q}\times\Bbb{Q}^\times$. The product is defined to be $$(a,r)(b,s)=(a+br,rs).$$ Observe that this is a group since $(0,1)(b,s)=(b,s)=(b,s)(0,1)$, and $(a,r)\left(\frac{-a}{r},\frac{1}{r}\right)=(0,1)$.
We regard $\Bbb{Q}$ as a subgroup of $G$ by $q\mapsto (q,1)$. Now if $A=\Bbb{Z}\subseteq \Bbb{Q}\subseteq G$, $g=(0,2)$, then we have $$(0,2)(n,1)\left(0,\frac{1}{2}\right)=(2n,2)\left(0,\frac{1}{2}\right)=(2n,1),$$ so $$(0,2)\Bbb{Z}\left(0,\frac{1}{2}\right)=2\Bbb{Z}.$$ This is therefore a counterexample, since $gAg^{-1}\subseteq A$, but $gAg^{-1}\ne A$.