For some group $G$ acting on a set $X$, if we consider the map:
$f: G \rightarrow \mathrm{Orb}(x)$ $ g \mapsto gx $
Then first we can tell that this is a surjection by definition of an orbit. So to prove injection, we can try and see if that for any two elements $g_1, g_2 \in G$, do we get that $g_1 x = g_2x$.
Applying $g_2^{-1}$ we get
$g_2^{-1}g_1 x = g_2^{-1}g_2 x \implies g_2^{-1}g_1 x = x$
Then applying $g_1^{-1}$ we get
$ g_1^{-1} g_2 x = g_1^{-1} g_1 x \implies g_1^{-1} g_2 x= x$
From these two, we see that $g_1 \in g_2 G_x$ and $g_2 \in g_1G_x$ and this happens iff $g_1Gx = g_2Gx$, hence proving an injection and therefore a bijection.
Is this correct?