Groupoids isomorphism Let $G, G'$ be two groups and $X=\{x,y\}$ be a set of two elements. Consider a groupoid $\mathcal{G}$ with objects from $X$ such that Hom$(x,x)=G$ and Hom$(y,y)=G'$.
Suppose Hom$(x,y) \neq \emptyset$, i.e. there is a morphism between $x$ and $y$. I think this is possible if and only if this morphism is an isomorphism between $G$ an $G'$ (in fact, every morphism in a groupoid must be invertible, moreover it must respect multiplication, so it is an isomorphism of these groups). Also, I know that any two such morphisms are conjugate.
I need to prove: "Given two groupoids $\mathcal{G}_1$ and $\mathcal{G}_2$ satisfying the conditions above (i.e. both have $X$ as set of objects and $G, G'$ as hom-sets Hom$(x,x)$ and Hom$(y,y)$), if there is a morphism from $x$ to $y$ in both groupoids, then $\mathcal{G}_1$ and $\mathcal{G}_2$ are isomorphic".
[Two groupoids are isomorphic if they are isomorphic as category, i.e. there is an "invertible" functor between the two]. 
It seems quite an obvious statement, but I cannot prove it. I start with a functor between the groupoids, say $F: \mathcal{G}_1 \to \mathcal{G}_2$. Then, what? Should I concretely construct such a functor and show it is an isomorphism? Or is there an "abstract nonsensical way" to do it?

You should just construct the functor.  The objects map to themselves as do the endomorphisms.  All you really have to decide is how the map $F\colon\hom_{\mathcal G_1}(x, y) \to \hom_{\mathcal G_2}(x, y)$ is going to work.
Pick $\phi_1 \in \hom_{\mathcal G_1}(x, y)$ and $\phi_2 \in \hom_{\mathcal G_2}(x, y)$.  What you need to do is show that every $f \in \hom_{\mathcal G_1}(x, y)$ can be written in the form $\phi_1f'$ where $f' \in G$, and similarly for $\mathcal G_2$.  Then define $F(\phi_1f') = \phi_2f'$.
You'll have to show that $F$ respects composition and is a bijection on that homset, but that shouldn't be hard.  
