Fix $n \geq 1$, and let $\mathbb S^n$ denote the sphere of radius one centred at zero in $\mathbb R^{n+1}$. Let $SO(n+1)$ denote the Lie group of $n+1$ by $n+1$ orthogonal matrices with determinant one. There is a natural smooth (left) action of $SO(n+1)$ on $\mathbb S^n$ via $A \cdot x := Ax$, where the expression on the right is matrix multiplication. For each $A \in SO(n+1)$, let $\varphi_A:\mathbb S^n \rightarrow \mathbb S^n$ denote the diffeomorphism $x \mapsto Ax$.
Now, let $g$ is a Riemannian metric on $\mathbb S^n$. Suppose $g$ is $SO(n+1)$-invariant. That is, suppose that for each $A \in SO(n+1)$, $\varphi_A$ is a Riemannian isometry. Then is $g$ a constant multiple of the round metric on $\mathbb S^n$?
Hint: if $g^\circ$ is the standard round metric, write $g(v,w) = g^\circ (Av,w)$ for a symmetric and positive-definite linear map $A$. Show that the eigenspaces of $A$ are ${\rm SO}(n+1)$-invariant and apply Schur's lemma (or, its proof) to conclude that $A$ must be a multiple of the identity.