A connected groupoid is equivalent to a groupoid with one object This is same as saying a connected groupoid is equivalent to a group. But I have no idea how to construct such a group, can't even get started.
Any help is appreciated.
 A: Using a different terminology, but basically similar reasoning as Dori:
Every category is equivalent to its skeleton
Every connected groupoid is a category
In a skeleton you fuse together all isomorphic objects, therefore the skeleton of a connected groupoid is a category with a single element and invertible morphisms (a group)
So every connected groupoid is equivalent to a group
A: Let $\mathcal{Gr}$ be your groupoid category. Take any object $X \in \mathcal{Gr}$. Consider the subcategory $G$ consisting of just the single object $X$ and with $\operatorname{Hom}_G(X,X) = \operatorname{Hom}_\mathcal{Gr}(X,X)$. The category $G$ is a group because every element of $\operatorname{Hom}_G(X,X)$ can be composed by any other element, they're all invertible because $\mathcal{Gr}$ is a groupoid, and there is an identity morphism $X \to X$. 
Now consider the inclusion functor $G \hookrightarrow \mathcal{Gr}$. This functor is fully faithful by construction since $\operatorname{Hom}_G(X,X) = \operatorname{Hom}_\mathcal{Gr}(X,X)$, and it is essentially surjective since $\mathcal{Gr}$ is is connected so for any $Y \in \mathcal{Gr}$, there exists a morphism $X \to Y$ which is necessarily an isomorphism since $\mathcal{Gr}$ is a groupoid. Therefore the inclusion is an equivalence of categories. 
