What is the symmetry of the pion triplet ($\pi^{-}, \pi^{0}, \pi^{+}$)? Under the entry "Isospin" in Wikipedia, it states:

The pions are assigned to the triplet (the spin-1, $\mathbf{3}$, or adjoint representation) of $SU(2)$

Why is the symmetry not $SU(3)$ since there are three particles? And in what circumstance do we have an $SU(3)$ symmetry?
 A: As per  urging by @rob, here is the concise answer:
Isospin SU(2) has a doublet representation, (u,d); a triplet representation, the 3 πs; an isoquartet representation, the 4 Δs; and so on...  You already know this from angular momentum, since, SU(2) ~ SO(3) is also the group of rotations/angular momentum, except here in isospace, an abstract notional space:
The spin doublets, spin 1/2, correspond to isodoublets here, u,d quarks. The spin triplets, spin 1, like 3-vectors, correspond to isotriplets, the pions. The spin quartets, spin 3/2, correspond to the four Δ baryons, etc... All SU(2) irreps are real (in a slightly technical sense... even the spinors).
Now, unlike SU(2), flavor SU(3) has a truly complex representation, 
a triple (u,d,s); a real octet representation; a complex decuplet, etc...
Now you consider a real triplet of pions, thus a real 3-vector. You know this vector transforms under SO(3) ~ SU(2), just like rotations of real vectors, so the group is the isospin SU(2) as stated. 
However, if it were a spinor, instead, a complex triplet, it would have to  transform under an SU(3): you could not restrict the number of independent transformations of its components to SO(3), and you'd be stuck with SU(3), eight independent transformation directions.  
This is what dictates SU(3) for a complex triplet of quarks, (u,d,s); 
even though, historically, the logic went backwards: the complex triplet was suggested by the fundamental representation of flavor SU(3), inferred by the real meson octet!


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*In case you were stumped by the complex $\pi^{\pm}\equiv (\pi^1\pm i\pi^2)/\sqrt{2}$, this is just the spherical vector rewriting of the Cartesian components $\pi^1,\pi^2$, so group theoretically the pion is still a real triplet.

