Let $ M $ be a compact connected manifold. The degree of symmetry of $ M $, denoted $ N(M) $, is the maximum of the dimensions of the isometry groups of all possible Riemannian structures on $ M $. See for example


<https://www.ams.org/journals/tran/1969-146-00/S0002-9947-1969-0250340-1/S0002-9947-1969-0250340-1.pdf>


Two metrics are considered to be equivalent if they are isometric up to a constant multiple. (see comment from Robert Bryant)


I'm interested in manifolds $ M $ for which there is a unique up to equivalence metric with isometry group of dimension $ N(M) $. My guess is that there is always such a unique metric for manifolds of the form $ G/H $ for $ G $ a compact connected simple Lie group and $ H $ a closed subgroup. Moreover I would imagine that this unique up to equivalence metric is just the pushforward of the unique up to scaling biinvariant metric on the compact connected simple Lie group $ G $.


I believe all spheres $ S^n, n \geq 2 $ have this property. And the unique maximum symmetry metric is the round metric.


What about the manifold $ M=\mathbb{C}P^n $ of real dimension $ 2n $? Is it the case that
$$
N(\mathbb{C}P^n) =n(n+2)
$$
And moreover is it true that every metric on $ \mathbb{C}P^n $ whose isometry group has maximum dimension must be equivalent to the Fubini-Study metric?


Edit: Ok so the guess about spaces $ G/H $ was a little ambitious and as Robert Bryant points out it is wrong. My new guess is that a unique up to equivalence metric exists for any irreducible compact symmetric space $ M $. (Edit: I originally left out "irreducible" which Robert Bryant pointed out in the comments makes this obviously false).


Ok I made this guess into a new a new question


[Unique maximum symmetry metric on irreducible compact symmetric space](https://mathoverflow.net/questions/433914/unique-maximum-symmetry-metric-on-irreducible-compact-symmetric-space)



Maximum symmetry metric on $ \mathbb{C}P^n $
There's an easy counterexample to your guess: Let $M^6 = \mathrm{SU}(3)/\mathbb{T}^2$, where $\mathbb{T}^2\subset\mathrm{SU}(3)$ is the maximal torus (for example, the diagonal subgroup). In that case, there is a 3-parameter family of non-isometric metrics on $M^6$ that are invariant under $\mathrm{SU}(3)$, so they are not unique up to a constant (or even non-constant) scalar factor.


I imagine that $M^6$ does not carry a metric whose isometry group has dimension greater than $8=\dim\mathrm{SU}(3)$, but I don't have a proof handy.


On the other hand, it is true that any Riemannian metric on $\mathbb{CP}^n$ whose isometry group has dimension at least $n(n{+}2)$ must be isometric to a constant scalar multiple of the Fubini-Study metric. Here is one argument:


Suppose that a connected, compact group $G$ acts effectively and smoothly on $\mathbb{CP}^n$. Then, by averaging, there exists a $G$-invariant metric $g$. Moreover, since $H^2\_{dR}(\mathbb{CP}^n,\mathbb{R})\simeq\mathbb{R}$, it follows from the Hodge Theorem that there is a $g$-harmonic $2$-form $\omega$ that represents a generator of $H^2\_{dR}(\mathbb{CP}^n,\mathbb{R})$, and it is unique up to constant multiples. Since $G$ is connected, it follows that it must leave $\omega$ fixed. Moreover, because of the structure of the cohomology ring of $\mathbb{CP}^n$, the top-degree form $\omega^n$ must represent a generator of $H^{2n}\_{dR}(\mathbb{CP}^n,\mathbb{R})$. In particular, $\omega^n$ does not vanish identically.


Thus, there is a point $p\in\mathbb{CP}^n$ such that $\omega\_p\in \Lambda^2(T^\*\_pM)$ is a 2-form of full rank. Consider the stabilizer $G\_p\subset G$ of $p$. Since $G$ acts by isometries and $\mathbb{CP}^n$ is connected, $G\_p$ injects into $\mathrm{O}(T\_pM)$ by identifying $g\in G\_p$ with $g'(p):T\_pM\to T\_pM$. Moreover, $G\_p$ leaves $\omega\_p$ fixed. Thus, $G\_p$ must lie inside a subgroup of $\mathrm{O}(T\_pM)$ that fixes a complex structure $J:T\_pM\to T\_pM$ and hence must have dimension at most $\dim \mathrm{U}(n) = n^2$. Now, we have
$$
\dim G = \dim G\_p + \dim G/G\_p = \dim G\_p + \dim G{\cdot}p \le n^2 + 2n = n(n{+}2).
$$
If equality holds, then $\dim G\_p = n^2$ and $\dim G{\cdot}p = 2n = \dim \mathbb{CP}^n$. Thus, the orbit $G{\cdot}p$ is both open and closed in $\mathbb{CP}^n$, so $G$ acts transitively on $\mathbb{CP}^n$. It follows that $\omega$ is everywhere of full rank and, after scaling $\omega$ so that it has comass 1, we have that $\omega(u,v) = g(Ju,v)$ for a unique almost-complex structure $J$ on $\mathbb{CP}^n$ that is preservd by $G\_p$, which has the same dimension as the connected group $\mathrm{U}(g\_p,J\_p)\simeq \mathrm{U}(n)$. Thus, $G\_p = \mathrm{U}(g\_p,J\_p)$. Since $G\_p$ contains $-I\in\mathrm{U}(g\_p,J\_p)$, it follows that there is an element of $G$ that fixes $p$ and reverses all $g$-geodesics through $p$. Since $G$ acts transitively on $\mathbb{CP}^n$, it follows that $(\mathbb{CP}^n,g)$ is a Riemannian symmetric space. Using the classification, it follows that $G\simeq \mathrm{SU}(n{+}1)/Z$ (where $Z\simeq\mathbb{Z}\_{n+1}$ is the center of $\mathrm{SU}(n{+}1)$) and that the metric $g$ is, up to isometry, a constant multiple of the standard Fubini-Study metric.


