Suppose we have a theory with a non-abelian symmetry group $G$ that is spontaneously broken to the subgroup $H\subset G$--this is a global symmetry, $not$ a gauge redundancy. Let $X^a$ be the generators of the unbroken subgroup $H$ and $Y^m$ be the remaining generators. For each broken generator $Y^m$, there is a corresponding Goldstone $\pi^m(x)$. My question is the following: if we integrate out all of the massive modes so that we are just left with a low-energy theory of the Goldstones corresponding to $G/H$, what do infinitesimal transformations of $\pi^a(x)$ look like under the $X^a$ and $Y^m$ generators?
For abelian symmetries we expect the Goldstones to have shift symmetries under broken generators, but that is clearly not possible for a generic non-abelian symmetry group. I expect something much more complicated happens but I can't quite see what.