# Is there a class of non-symmetric spaces $G/SO(n)$ such that $Hol(G/SO(n))=SO(n)$?

Is there a certain class (or unique construction of a set) of simply connected non-symmetric spaces $G/SO(n)$ such that the Riemannian holonomy is $Hol(G/SO(n))=SO(n)$? (That is, is there a way to write $G$ as a unique Lie group or a class of groups to guarantee this, and must there be infinitely many such non-symmetric spaces?)

Can we enumerate all such possible candidates or classify them (or is this even possible)? If this is plausible, it would be interesting to see how it extends to other quotients like $G/SU(n)$, $G/Sp(n)$, etc. I would like to enumerate all such spaces, but there is still the question of whether there are infinitely-many or finitely-many such (unique) spaces. Since $G/SO(n)$ is simply connected by assumption, the holonomy and isotropy group are equal so $Hol(G/SO(n))=SO(n)=\{g\in G:gx=x\}$ for some $x\in G/SO(n)$.

Cross-posted on MO.