Let $M$ be a closed connected smooth manifold. We define the degree of symmetry of $M$ by $N(M):=\sup_\limits{g}\mathrm{dim}\,\mathrm{Isom}(M,g)$, where $g$ is a smooth Riemannian metric on $M$ and $\mathrm{Isom}$ is the isometry group of the Riemannian manifold $(M,g)$.

The torus $T^n$ does not admit a Riemannian metric with positive scalar curvature and has $N(T^n)\neq 0$.

Is there a closed connected manifold $M$ with $N(M)=0$ such that $M$ admits a metric with positive scalar curvature? That is, does admitting a metric with positive scalar curvature imply its degree of symmetry is nonzero?