Let $G_1, G_2$ be two groups that act on a set $S$ on the left, such that for all $g \in G_1$ there's $g' \in G_2$ such that $g\cdot s = g'\cdot s$ for all $s \in S$.
Define $h : G_1 \to G_2$, $h(g) = g'$ as defined above. Then $h(ab) = (ab)'$. Well $(ab)\cdot s = (ab)'\cdot s \ \forall s \implies \text{ it equals } a(b\cdot s) = a'(b'\cdot s)$. But I can't conclude that $(ab)' = a'b'$.
How are the two groups related?
Thanks @ThomasAndrews. "If the group actions are faithful then $G_1 \approx G_2$."
Let $h$ be well-defined by choosing any such $g'$ for each $g \in G_1$. From the above, as $(ab)' \cdot s = a'(b' \cdot s)$, we have that $b'^{-1} a'^{-1} (ab)' = 1$, so that equality follows. That means if $G_2 \times S \to S$ is faithful then $h$ is a homom. $h$ is injective as $g'\cdot s = h' \cdot s \ \forall s$, then by faithfullness of $G_2$ again $g' = h'$.
Intuitively, if we have the similar requirement and a map $h' : G_2 \to G_1$, then $h$ becomes an isomorphism.
If the group actions are faithful, then Thomas Andrews's comment answers the question. For general group actions, the situation is as follows. In each of your two groups $G_i$, the subset $H_i$ of elements that act as the identity on $S$ (i.e., map every element of $S$ to itself) is a normal subgroup. The two quotient groups $G_i/H_i$ are isomorphic.