Let $M$ be a closed $n$-dimensional Riemannian manifold with positive sectional curvature. Let $G$ be a close subgroup of isometry group ${\rm Iso}(M)$. Suppose the action of $G$ on $M$ is not transitive, hence $M/G$ has dimension at least $1$.
By a theorem of Grove and Searle the symmetry rank $${\rm symran}(M)\le [\frac{n+1}{2}]$$ I am wondering is there any upper bound for the dimension of $G$ mentioned above?
Forgetting positive curvature, if $\dim M^n/G=k$ then by looking at the transitive action of $G$ on the principal orbit one gets a trivial bound $\dim G\le \dim O(n-k)=\frac{(n-k+1)(n-k)}{2}$. This bound is realized for $k=1$ on a round $S^n$ and $G=O(n-1)$. As the sphere is positively curved this bound is sharp.
Addressing the comment below, the assumption of $M/G$ being a manifold is not a natural one in this context. It hardly ever happens when $M$ has positive curvature. In particular, by a result of Wilking (Lemma 5 in "Positively curved manifolds with symmetry") based on his connectedness principle, if the principal isotropy group $H$ is not trivial and $M/G\ne pt$ then $M/G$ has a boundary. If $H=1$ and $M/G$ is a smooth manifold then the $G$-action is free and hence $rank G\le 1$ by Berger's vanishing theorem.
I should add that there is large literature on the subject of isometric group actions on positively curved manifolds (mostly by Wilking, Grove, Ziller, Searle and Rong) and I suggest you study it if you want to pursue these kind of questions. A good place to start is a survey by Wilking "Nonnegatively and positively curved manifolds".
