Consider the $n$-sphere $$ S^n = \{x\in\mathbb{R}^{n+1}: 1 - \sum\_{k=1}^{n+1} x\_k^2 = 0\}, $$ and let $g\_1$ be the induced metric. Given $\lambda\in\mathbb{R}^{n+1}\_{>0}$, we have the ellipsoid
$$
E\_\lambda \longrightarrow \{x\in\mathbb{R}^{n+1}: 1 - \sum\_{k=1}^{n+1} \lambda\_k x\_k^2 = 0\}.
$$
Let $g\_\lambda$ be the induced metric on $E\_\lambda$. A basic fact is that $S^n$ and $E\_\lambda$ are diffeomorphic, with the map $\psi\_\lambda : S^n \longrightarrow E\_\lambda$ given by $$\psi\_\lambda(x\_1,\ldots, x\_{n+1}) = \left(\frac{x\_1}{\sqrt{\lambda\_1}},\ldots, \frac{x\_{n+1}}{\sqrt{\lambda\_{n+1}}}\right).$$


Now let $H=\text{diff}(S^n)$ be the diffeomorphism group of $S^n$, a group whose definition is independent of the choice of metric on the sphere. $H$ acts (on the right) on the space of Riemannian metrics on $S^n$ via pullback: if $g$ is a metric on $S^n$, then $\varphi^\ast g$ is another Riemannian metric. I am interested in the orbit of $g\_1$ under the action of $H$ on the space of Riemannian metrics.



> 
> **Question:** Does there exist $\varphi\in\text{diff}(S^n)$ such that $\varphi^\ast g\_1$ is equal to $\psi\_\lambda^\ast g\_\lambda$?
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Orbits of Metrics under the Action of the Diffeomorphism Group
Answer: Not unless all of the $\lambda\_i >0$ are equal to $1$ (or else, when $n=1$ and $\lambda\_1$ and $\lambda\_2$ are chosen so that the length of the curve $E\_\lambda$ is $2\pi$; the condition for this is given by evaluating an elliptic integral). 


For $n>1$, this follows from a calculation of the eigenvalues of the Riemann curvature tensor of $g\_\lambda$ which is a diffeomorphism invariant.


