Here $G$-sets denote the category of sets which have a left $G$-action. So the question is whether a functor $F \colon \text{$G$-sets} \to \text{$H$-sets}$ implies that we have an isomorphism of groups $\phi \colon G \to H$.
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The category of left $G$-sets is equivalent to the category of left $H$-sets if and only if $G$ is isomorphic to $H$.
Indeed, consider $G$ as a left $G$-set. It has the property that $\mathrm{Hom}_G (G, -) : G \textbf{-Set} \to \mathbf{Set}$ preserves all colimits. Moreover, up to isomorphism, $G$ is the unique such left $G$-set, i.e. if $X$ is a left $G$-set such that $\mathrm{Hom}_G (X, -) : G \textbf{-Set} \to \mathbf{Set}$ preserves all colimits, then $X \cong G$. In addition, $G$ can be recovered as a group: $G \cong \mathrm{Aut} (X)$. Thus $G$ (as an abstract group) can be recovered from $G\textbf{-Set}$ (as an abstract category).
The above arguments are very specific to this situation. For example, if you replace $\mathbf{Set}$ with $\mathbf{Ab}$ then you can have non-trivial equivalences of categories of representations. This is the phenomenon of Morita equivalence.
Zhen Lin
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1Sure it's not $G^{\operatorname{op}} \cong \operatorname{Aut}X $ ? – darij grinberg Jul 27 '15 at 09:49
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3Well, $G \cong G^\mathrm{op}$, so there's no real difference. But yes, in some sense it's really $G^\mathrm{op}$. – Zhen Lin Jul 27 '15 at 10:50
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Ah, I was stupid. – darij grinberg Jul 27 '15 at 10:59
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1In more abstract terms, you recover the fiber functor and then reconstruct the group as usual as the automorphism group of the fiber functor. – Martin Brandenburg Aug 03 '15 at 08:55
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Indeed. In the case of linear representations, the fibre functor is indispensable. – Zhen Lin Aug 03 '15 at 09:37