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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.

I'm interested in manifolds $ M $ for which there is a unique up to equivalence metric with isometry group of dimension $ N(M) $.

Is it true that for any irreducible compact symmetric space $ M=G/H $ there is a unique up to equivalence metric with maximum symmetry? If so, 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 $.

Some examples supporting the claim:

I believe all spheres $ S^n, n \geq 2 $ have this property. And the unique maximum symmetry metric is the round metric, which is indeed the pushforward of the biinvariant metric on $ SO_{n+1} $.

Also all complex projective spaces $ \mathbb{C}P^n $ have this property. And the unique maximum symmetry metric is the Fubini-Study metric, which is indeed the pushforward of the biinvariant metric on $ SU_{n+1} $. See

Maximum symmetry metric on $ \mathbb{C}P^n $

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    $\begingroup$ At least the sphere case can be settled: fix a point and let it's stabiliser act on the tangent space. In the sphere case the stabiliser is SO(n) and there is, up to scaling) only one inner product being invariant under SO(n). The metric at this point fixes it at all other points, as the metric is transported via the isometry group. $\endgroup$
    – user473423
    Commented Nov 5, 2022 at 8:31
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    $\begingroup$ This doesn't address your question, but it might be worth pointing out: in the non-compact type, the symmetric metric can be viewed as a left-invariant metric on certain solvable Lie group $S$ thanks to the Iwasawa decomposition. Among all left-invariant metrics on $S$, the symmetric one $g_{sym}$ is the one with the largest isometry group in an even stronger sense: for any other left-invariant $g$, there exists an automorphism $\varphi \in Aut(S)$ such that $\varphi^*g$ has isometry group contained in that of $g_{sym}$ (this is due to Jablonski and Gordon, JDG 2019). $\endgroup$
    – Ramiro Lafuente
    Commented Nov 24, 2022 at 0:13
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    $\begingroup$ Also, the result mentioned in the previous comment can also be proved by Ricci flow: sciencedirect.com/science/article/pii/S000187081930297X $\endgroup$
    – Ramiro Lafuente
    Commented Nov 24, 2022 at 0:14

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