Is a connected groupoid uniquely determined by its objects and fundamental group? A small connected groupoid is such a category $G$ that:

*

*$\text{Ob}(G)$ is a nonempty set

*$\text{Hom}_G(a,b)$ is a nonempty set for all $a,b\in\text{Ob}(G)$

*$f$ is invertible for each $f\in\text{Hom}_G(a,b)$
One may show that $\text{Hom}_G(a,a)$ is a group and all such groups are isomorphic. This gives us the notion of fundamental group.
Can we prove that $G$ is uniquely (up to isomorphism) determined by $\text{Ob}(G)$ and its fundamental group?
 A: Under the axiom of choice, yes. Suppose we have two groupoids $X, Y$ with an isomorphism $\varphi: \pi_1(X, x_0) \simeq \pi_1(Y, y_0) $ and a bijection $\psi : \operatorname{Ob}X \to \operatorname{Ob}Y$. Suppose WLOG that $\psi$ maps $x_0$ to $y_0$. Here $\pi_1(X, x_0) := \operatorname{Hom}_X(x_0, x_0) $ is the fundamental group of $X$ at $x_0$.
Now we will choose a "star" for each of the two groupoids. For any $x \in X$ choose an isomorphism $f_x : x_0 \to x$, and analogously for any $y \in Y$ an isomorphism $g_y: y_0 \to y$ .
Define a functor $\psi: X \to Y$ in the following way. On objects, it correspond to the given bijection $\psi$. The star is mapped to the star, that is $ \psi(f_x) = g_{\psi(x) }$. The fundamental group is mapped to the fundamental group, that is $\psi(g) = \phi(g)$ for $g:x_0 \to x_0$.
Finally, note that a general morphism $h: a \to b$ in $X$ can be "pulled back" to the fundamental group by pre and post composing with the star. Formally, the element $ f_b^{-1} h f_a $ is in $ \pi_1(X, x_0) $. This determines where the element should be mapped from our functor, that is
$$ \psi(h) = \psi( f_b f_b^{-1} h f_a f_{a^{-1}} ) = \psi (f_b) \psi(f_b^{-1} h f_a) \psi(f_a) ^{-1} = g_{\psi(b) } \varphi( f_b^{-1} h f_a) g_{\psi(a) }^{-1} $$
With this general definition, it's easy to see that the identity of an element $a$ is mapped to $1_{\psi(a) }$. I will leave functoriality as an exercise: you will see that all that have to cancel actually cancels! Also, note that the functor is bijective on both objects and hom-sets, so it is an isomorphism of categories. Actually, you can produce the inverse by reproducing the same construction starting with $\psi^{-1}$ on objects and $\varphi^{-1}$ on the fundamental group.
A: Let $G$ be a connected groupoid. In particular, it is non-empty, so choose some object $a$. I claim that $G$ is isomorphic to the "trivial" groupoid $\mathrm{Ob}(G) \times \mathrm{Aut}(a)$ (where we regard $\mathrm{Ob}(G)$ as a discrete groupoid and the group $\mathrm{Aut}(a)$ as a groupoid with the single object $a$). In particular, $G$ is determined by the set $\mathrm{Ob}(G)$ and the group $\mathrm{Aut}(a)$.
In fact, choose isomorphisms $\gamma_b : a \to b$ for every $b \in \mathrm{Ob}(G)$. Define a functor $F : \mathrm{Ob}(G) \times \mathrm{Aut}(a) \to G$ on objects by $(b,a) \mapsto b$ and on morphisms by $(\mathrm{id} : b \to b,f : a \to a) \mapsto \gamma_b \circ f \circ \gamma_b^{-1}$. It is easy to check $F$ is indeed a functor. Also, $F$ is a bijection on objects (clear) and on morphisms since the conjugation $\mathrm{Aut}(a) \to \mathrm{Aut}(b)$ with $\gamma_a$ is inverse to the conjugation with $\gamma_a^{-1}$. Hence, $F$ is an isomorphism of categories.
Notice that this isomorphism is not canonical. It involves the choice of the isomorphisms $\gamma_b$.
