Sometimes when you define a group using an arbitrary choice of object and then show the choice of object doesn't matter, you could have defined a groupoid without making an arbitrary choice.

For example, to define the fundamental group $\pi_1(X,x)$ of a path-connected space $X$ we need to choose a basepoint $x \in X$, but then we can show we get isomorphic groups no matter what basepoint we choose, with an isomorphism given by a homotopy class of paths between the basepoints. To avoid this maneuver we can work with the fundamental groupoid of $X$, whose objects are points of $X$ and whose morphisms are homotopy classes of paths. If $X$ is path-connected all objects in this groupoid are isomorphic, and thus the automorphism groups of all objects are isomorphic. The automorphism group of $x$ is just $\pi_1(X,x)$. The fundamental groupoid is thus equivalent, as a category, to the one-object groupoid corresponding to the group $\pi_1(X,x)$. But the advantage of the fundamental groupoid is that we can define it without choosing a basepoint, and it makes sense and works well even when $X$ is not path-connected.

Similarly, I think we can define the **Weyl groupoid** of a compact semisimple Lie group $G$ in a way that gives a groupoid equivalent to the usual Weyl group, but doesn't require a choice of maximal torus. The idea should go like this. The objects of the Weyl groupoid are maximal tori. A morphism $f : T \to T'$ in the Weyl groupoid is a Lie group isomorphism of the form

$$ t \mapsto g t g^{-1} \textrm{ for all } t \in T $$

for some $g \in G$. If I did this right, the automorphism group of any object $T$ in the Weyl groupoid is the usual Weyl group

$$ W_G(T) = N_G(T) / T ,$$

that is, the normalizer of $T \subset G$ modulo the centralizer of $T \subset G$, which is $T$ itself. If this is true, the Weyl groupoid will be equivalent, as a groupoid, to the usual Weyl group $W_G(T)$ for any maximal torus $T$.