Consider $(S^{n},g_{round})$ be the standard unit sphere in $\mathbb{R}^{n+1}$ and $g_{round}$ be the metric on $S^{n}$ induced by the standard metric from $\mathbb{R}^{n}$. The task is to show that the isometry group of $(S^{n},g_{round})$ consists of $O|_{S^{n}}$, where $O\in \mathcal{O}(n)$ (orthogonal group).
So far I have the following: Let $f : S^{n} \rightarrow S^{n}$ be an isometry (meaning $f^{*}g_{round} = g_{round}$). Extend this to a contunous map $\tilde{f} : \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$ by setting $\tilde{f}(x) = ||x||f(x/||x||)$ and $\tilde{f}(0)=0$. Now I am trying to show that $\tilde{f}$ preserves distances. By that I mean that $||\tilde{f}(x) - \tilde{f}(y)|| = ||x-y||$ for all $x,y \in \mathbb{R}^{n+1}$. If I can show this then I can deduce that $\tilde{f}(x)=Ox$ for some $O \in \mathcal{O}(n)$.
Question: How do I show that $\tilde{f}$ preserves distances? For a hint I would be very thankful.
Greetings, Phil