Why, conceptually, isn't the 2-category of connected groupoids equivalent to the 2-category of groups?

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It is well-known that every connected groupoid is equivalent to a group. However the 2-category of connected groupoids is not equivalent to the 2-category obtained trivially from the 1-category of groups. This clashes with my intuition: the first statement is intuitively saying that a connected groupoid is the same thing as a group, and the second is intuitively saying that the collection of all connected groupoids somehow behaves different than the collection of all groups. What causes this mismatch? Is thinking of equivalent objects in math as "the same" bad, and if so, how should we think about equivalent objects?

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the first statement is intuitively saying that a connected groupoid is the same thing as a group

No it's not. It's saying that isomorphism classes of connected groupoids can be identified with isomorphism classes of groups. But an equivalence of categories or $2$-categories needs to consider morphisms (and $2$-morphisms), not just objects. You can show that the $2$-category of connected groupoids is equivalent to the $2$-category whose

  • objects are groups $G$,
  • morphisms are group homomorphisms $f : G \to H$, and
  • $2$-morphisms between two parallel group homomorphisms $f, g : G \to H$ are elements $h \in H$ such that $hf(x) h^{-1} = g(x)$ for all $x \in G$.

It's the $2$-morphisms here that make this $2$-category not equivalent to the category of groups. For example, $\text{Hom}(\mathbb{Z}, G)$ is the groupoid whose objects are the elements of $G$ and whose morphisms are conjugacies between these elements, and $\text{Hom}(\text{id}_G, \text{id}_G)$ (the $2$-morphisms from the identity homomorphism $\text{id}_G : G \to G$ to itself) is the center $Z(G)$.

What is true is that the $2$-category of pointed connected groupoids is equivalent to the category of groups; the addition of basepoints removes the interesting $2$-morphisms above and allows us to write down a canonical functor to $\text{Grp}$ given by taking the automorphism group of the basepoint. Without the basepoint, to actually write down a group starting from a connected groupoid you have to pick a basepoint, and there isn't a canonical one.

This distinction between connected things and pointed connected things is much more significant than it might appear at first glance and rears its head in many other ways. For example, you can define what it means for a group $G$ to act on a connected groupoid $X$, and you can find examples of such actions that don't arise from actions of $G$ on the group $\pi_1(X, x)$. This is because actions arising from an action on $\pi_1$ need to have a fixed point and they don't always; see e.g. this MO answer. (I am being a bit sloppy here; there are two possible definitions of an action and also two possible definitions of a fixed point depending on $2$-categorical you're willing to be.)

Here's another example: recall that the absolute Galois group of a field $K$ is the Galois group $\text{Gal}(\bar{K}/K)$. It behaves in many ways like a fundamental group (of the affine scheme $\text{Spec } K$), and is generalized by the etale fundamental group. It would be nice if it were a (contravariant) functor. But it isn't! The reason is that the algebraic closure is not actually unique, only unique up to (non-unique) isomorphism, and to define the absolute Galois group you have to pick an algebraic closure, which is the analogue in this setting of picking a basepoint. The notation is a bit deceptive here because we write the algebraic closure $\bar{K}$ as if it were a functor but it isn't one either.

The functorial construction is not a group but a connected groupoid, namely the connected groupoid of all algebraic closures of $K$. And now the functor is easy to write down: given a morphism $K \to L$ of fields, every algebraic closure $\bar{L}$ of $L$ canonically induces an algebraic closure of $K$, namely the algebraic closure of $K$ in $\bar{L}$.