TITLE: "Maximal symmetry" metric for a manifold?
QUESTION [10 upvotes]: So this is my question : 
Let $M$ be a smooth manifold. With any riemaniann metric $g$ on $M$ comes an isometry group $I(g)$. Intuition (well at least mine, which may be flawed...) suggests that there should be a metric $g$, not necessarily unique, that gives maximal symmetry to $M$ in the following sense : 
if $g'$ is any riemannian metric on $M$ then there is an injective morphism from $I(g')$ to $I(g)$.
The example I have in mind is the sphere $S^2$ : it seems clear that the usual metric gives maximal symmetry and that any other metric would either give the same group of symmetry or one strictly smaller.
Does anyone know if this is true or not ? If so can you give me the argument or a reference ? 
Thanks !

REPLY [5 votes]: As others have pointed out, the answer is definitely no. Here's a short sketch of another proof: For any dimension $n$, the metrics (on connected manifolds) with the isometry groups of the largest dimension are precisely the space forms, that is, the metrics of constant sectional curvature. Up to rescaling and passing to the universal cover (neither of which can increase the dimension of the isometry groups), these spaces are precisely

the sphere $\mathbb{S}^n$ with the round metric
Euclidean space $\mathbb{R}^n$ with the flat metric
hyperbolic space $\mathbb{H}^n$ with the hyperbolic metric.

The respective isometry groups are $SO(n + 1)$, $SO(n) \rtimes \mathbb{R}^n$, and $SO(n, 1)$, each of which has dimension $\frac{1}{2} n (n + 1)$.
Now, $\mathbb{R}^n$ and $\mathbb{H}^n$ are diffeomorphic to one another but there is no group of dimension $\frac{1}{2} n (n + 1)$ into which their respective isometry groups both inject.