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.

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 Wiemeler https://arxiv.org/pdf/1108.5327.pdf.

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 $0$.

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.