Groups isomorph iff categories equivalent

Working through some category theory (finally!) I'm wondering if equivalence of categories is already strong enough to invoke an isomorphism between groups.

So, more formally:

Given two Groups $G$, $H$, if seen as categories $\mathcal{G}, \mathcal{H}$, does the following hold:

$$\mathcal{G}\simeq\mathcal{H} \Rightarrow G \cong H$$

Ideas

Pro

If we allow $\mathcal{G} \cong \mathcal{H}$ then the given functors are already the needed homomorphisms. Given only equivalence, one has the following properties, given the functors (i.e. group homomorphisms on arrows) $\phi : G \to H$ and $\psi : H \to G$:

$$(1)\;\;\; \forall g \in G: \psi(\phi(g))\cdot\eta = \eta \cdot g$$ $$(2)\;\;\; \forall h \in H: \epsilon\cdot\phi(\psi (h)) = h\cdot \epsilon$$

where $\eta, \epsilon$ are fixed elements from $G$ and $H$ resp. given by the natural transformations.

There might be some way to come up with an isomorphism using $\phi, \psi, \eta$ and $\epsilon$ but I didn't succeed. Probably, since my group theory classes are somewhat in the far past, I cannot see a natural isomorphism here.

Contra

Other way round: I've tried to come up with some counter-examples. Again, being more into group theory would have helped; somehow I didn't managed that either (although considering trivial examples).

Stating a counterexample would be fine for me; best would be some insight, since I'm trying to understand what it does mean for categories to be equivalent. Thanks in advance! : )

• It turns out that a category $\mathcal{C}$ is equivalent to $\mathcal{G}$ if and only if $\mathcal{C}$ is a connected groupoid (i.e. every hom-set is nonempty and all morphisms are invertible) where $\hom(X,X) \cong G$ for any (and thus all) objects $X$ of $\mathcal{C}$.
– user14972
Sep 14 '16 at 22:09

Another characterization of an equivalence $T: \mathcal{C} \to \mathcal{D}$ is that it is an "essentially surjective fully faithful functor", where "essential surjectivity" means that for every object $d\in \mathcal{D}$ there is $c\in \mathcal{C}$ such that $T(c)\cong d$.
Since each of $\mathcal{G}$ and $\mathcal{H}$ have a single object, an equivalence is the same thing as a fully faithful functor.
A fully faithful functor $T: \mathcal{G} \to \mathcal{H}$ is characterized entirely by the induced map $\hom_{\mathcal{G}}(\ast,\ast) \to \hom_{\mathcal{H}}(\ast,\ast)$. In general this induced map is a monoid homomorphism (since $\hom_{\mathcal{C}}(c,c)$ is a monoid under composition and functors preserve composition). The hypothesis of full and faithful means that this induced map is a bijection, and vice-a-versa.