Maximum symmetry metric on $ \mathbb{C}P^n $ 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
I'm interested in manifolds $ M $ for which there is a unique, up to scaling, metric with isometry group of dimension $ N(M) $. This number is always bounded by
$$
N(M) \leq \frac{n(n+1)}{2}
$$
where $ n $ is the dimension of the manifold $ M $.
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
$$
N(\mathbb{C}P^n) =n(n+2)
$$
an isometry group dimension which is achieved by the Fubini-Study metric?
And moreover is it true that every metric on $ \mathbb{C}P^n $ whose isometry group has maximum dimension must be a scalar multiple of the Fubini-Study metric?
 A: I cross-posted to MO and Robert Bryant gave this lovely answer
https://mathoverflow.net/questions/433847/maximum-symmetry-metric-on-mathbbcpn
which makes a clever argument that a complex structure must be preserved and then uses that to establish the bound
$$
dim(G)= dim(G_p)+dim(G/G_p)\leq dim(U_n) + dim(\mathbb{CP}^n)=n^2+2n
$$
(since the largest group $ G_p $ that can preserve a complex structure on a $ 2n $ dimensional manifold is $ U_n $) where $ G $ is any compact connected group acting effectively and smoothly. This establishes the bound
$$
N(\mathbb{CP}_n)\leq n^2+2n=(n+1)^2-1=dim(PU_{n+1})
$$
The Fubini-Study metric saturates this bound so of course we have
$$
N(\mathbb{CP}_n)= (n+1)^2-1
$$
From there he argues that any metric $ g $ whose isometry group $ G $ saturates this bound must actually make $ \mathbb{CP}^n $ into a symmetric space. Then he appeals to the classification of symmetric spaces to conclude that $ G=PU_{n+1} $ and the metric $ g $ must be a scalar multiple of Fubini-Study.
I believe a similar argument works for quaternionic projective space $ \mathbb{HP}^n $:
Suppose that a connected, compact group $G$ acts effectively and smoothly on $\mathbb{HP}^n$.  Then, by averaging, there exists a $G$-invariant metric $g$.  Moreover, since $H^2_{dR}(\mathbb{HP}^n,\mathbb{R})\simeq\mathbb{R}$, it follows from the Hodge Theorem that there is a $g$-harmonic $2$-form $\omega$ that represents a generator of $H^2_{dR}(\mathbb{HP}^n,\mathbb{R})$, 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 $\mathbb{HP}^n$, the top-degree form $\omega^n$ must represent a generator of $H^{2n}_{dR}(\mathbb{HP}^n,\mathbb{R})$.  In particular, $\omega^n$ does not vanish identically.
Thus, there is a point $p\in\mathbb{HP}^n$ 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 $\mathbb{HP}^n$ 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 quaternionic structure $J:T_pM\to T_pM$ and hence must have dimension at most $\dim (\mathrm{Sp}(n) \times Sp_1)= n(2n+1)+3$.  Now, we have
$$
\dim G = \dim G_p + \dim G/G_p = \dim G_p + \dim G{\cdot}p =dim(Sp_n\times Sp_1) + dim(\mathbb{HP}^n)
$$
Note that
$$
dim(Sp_n\times Sp_1) + dim(\mathbb{HP}^n)=n(2n+1)+3 + 4n = (n+1)(2(n+1)+1)= dim(Sp_{n+1})
$$
If equality holds, then $\dim G_p = n(2n+1)+3$ and $\dim G{\cdot}p = 4n = \dim \mathbb{HP}^n$.  Thus, the orbit $G{\cdot}p$ is both open and closed in $\mathbb{HP}^n$, so $G$ acts transitively on $\mathbb{HP}^n$.  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-quaternionic structure $J$ on $\mathbb{HP}^n$ that is preserved by $G_p$, which has the same dimension as the connected group $\mathrm{Sp}(g_p,J_p)\simeq \mathrm{Sp}(n)$.  Thus, $G_p = \mathrm{Sp}(g_p,J_p)$.  Since $G_p$ contains $-I\in\mathrm{Sp}(g_p,J_p)$, it follows that there is an element of $G$ that fixes $p$ and reverses all $g$-geodesics through $p$.  Since $G$ acts transitively on $\mathbb{HP}^n$, it follows that $(\mathbb{HP}^n,g)$ is a Riemannian symmetric space.  Using the classification, it follows that $G\simeq \mathrm{Sp}(n{+}1)/Z$ (where $Z\simeq\mathbb{Z}_{2}$ is the center of $\mathrm{Sp}(n{+}1)$) and that the metric $g$ is, up to isometry, a constant multiple of the standard metric on $ \mathbb{HP}^{n+1} $.
