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Let $M$ be a closed connected smooth manifold, then 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 $N(T^n)\neq 0$.

Whether there exists $M$ with $N(M)=0$ such that $M$ admits a metric with positive scalar curvature? That is, whether admitting a metric with positive scalar curvature implies its degree of symmetry is nonzero?

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    $\begingroup$ You can replace the supremum in the definition of $N(M)$ with a maximum because $\dim\operatorname{Isom}(M, g) \leq \frac{1}{2}n(n+1)$. $\endgroup$ Jul 29 at 15:44
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It seems that there are examples. By a theorem of Gromov and Lawson every simply connected manifold of dimension $n \geq 5$ which is not spin admits a metric of positive scalar curvature.

There are many examples of simply connected, non-spin, closed $6$-manifolds which cannot admit a smooth circle action, constructed by Puppe. Theorem 7 of https://arxiv.org/pdf/math/0606714.pdf.

Then, since the isomotetry group of a closed manifold is a compact Lie group, if $N(M)>0$ then taking a maximal torus gives a non-trivial circle action, which contradicts the above. So every metric has isometry group of dimension $0$.

Edit: A specific example would be a quartic $3$-fold $X \subset \mathbb{CP}^4$. It admits a metric with positive Ricci curvature (since it is Fano), or alternatively since it is not spin we can apply Gromov-Lawson. It does not admit any smooth circle action due to a Theorem of Dessai and Wiemler https://arxiv.org/pdf/1108.5327.pdf.

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