Let $(M,g)$ be a Riemannian manifold of dimension $n$. Let $\text{conf}(M,g)$ denote the conformal group, i.e. the subgroup of diffeomorphisms of $M$ that acts by conformal transformations relative to the metric $g$. Set $$ \mathcal{O}_g = \{\varphi^\ast g : \varphi\in \text{conf}(M,g)\}. $$

Let's call $\mathcal{O}_g$ the conformal orbit of $g$. Next, we consider the conformal class of $g$: $$ [g] = \{e^{2\omega}g : \omega\in C^\infty(M)\}.$$

Generally, the conformal orbit and the conformal class are different objects. For example, if $(M,g)$ is a Riemannian manifold such that $\dim \text{conf}(M,g) < {n+2\choose 2}$, then the conformal Lichnerowicz theorem implies the existence of a metric $\tilde g\in [g]$ such that $\text{conf}(M,\tilde g)$ is equal to the isometry group of $\tilde g$. It follows that $\mathcal{O}_{\tilde g}$ is a single point and hence much smaller than $[\tilde g]$.

Question: Let $(M,g)$ be a Riemannian manifold such that $\dim \text{conf}(M,g) = {n+2\choose 2}$. Is $\mathcal{O}_g = [g]$?

If so, then this would apply to the standard $n$-sphere, since the identity component of its conformal group is isomorphic to $SO_0(n+1,1)$, a Lie group of dimension ${n+2\choose 2}$.

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    $\begingroup$ Isn't the first space finite dimensional for $n > 2$ and the second not? $\endgroup$ – Paul Reynolds Aug 22 '15 at 21:49

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