Isometries of $\mathbb{S}^n$

This isn't too difficult to show directly. Any isometry $f\colon\mathbb{S}^n\to\mathbb{S}^n$ extends easily to a map $g\colon\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$, (writing $\hat x\equiv x/\Vert x\Vert\in\mathbb{S}^n$ for $x\in\mathbb{R}^{n+1}\setminus\{0\}$) $$ g(x)=\begin{cases} \Vert x\Vert f(\hat x),&\textrm{if }x\not=0,\cr 0,&\textrm{if }x=0. \end{cases} $$ It is clear that $f$ is the restriction of $g$ to $\mathbb{S}^n$, so all that needs to be done is to show that $g$ is an isometry. In particular, $$ \Vert g(x)-g(y)\Vert=\Vert x-y\Vert\qquad{\rm(1)} $$ for $x,y\in\mathbb{R}^{n+1}$. Also, $\Vert g(x)-g(0)\Vert=\Vert x\Vert\Vert f(\hat x)\Vert=\Vert x\Vert=\Vert x-0\Vert$, so (1) only needs to be shown for nonzero $x,y$. However, the distance between $x,y$ can be written purely in terms of $\Vert x\Vert,\Vert y\Vert$ and $\hat x\cdot\hat y$, $$ \begin{align} \Vert x-y\Vert^2&=\Vert x\Vert^2+\Vert y\Vert^2-2x\cdot y\cr &=\Vert x\Vert^2+\Vert y\Vert^2-2\Vert x\Vert\Vert y\Vert\hat x\cdot\hat y. \end{align}\qquad(2) $$ Now, $g$ preserves the norm of any $x\in\mathbb{R}^{n+1}$. Also, $\hat x\cdot\hat y$ is just the cosine of the distance from $\hat x$ to $\hat y$ along the sphere. So, this is preserved by $f$ and, hence, $\widehat{g(x)}\cdot\widehat{g(y)}=f(\hat x)\cdot f(\hat y)=\hat x\cdot\hat y$. Applying (2) to both $\Vert x-y\Vert$ and $\Vert g(x)-g(y)\Vert$ shows that (1) holds and $g$ is distance preserving.

As the (Riemannian) metric is determined by the distance between points (note: if $\gamma$ is a smooth curve in a manifold $X$, $g(\dot\gamma(t),\dot\gamma(t))=\lim_{h\searrow0}h^{-1}d(\gamma(t),\gamma(t+h))$), then as long as it is known that $g$ is smooth, the argument above implies that $g$ is an isometry in the Riemannian sense. It is known that any distance preserving invertible map between $n$-dimensional manifolds is smooth. If you don't want to use such a result, you can instead use the fact that any distance (and origin) preserving map $\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ is linear (e.g., see this question), hence smooth.

Denote the isometry f, and consider the matrix with columns $f(e_1), f(e_2), f(e_3), \ldots$, where ${e_i}$ form an orthnormal basis of $\mathbb{R}^{n+1}$, also conveniently lying on the sphere. This matrix is orthonormal and so defines an isometry of $\mathbb{R}^{n+1}$

The intuition here is that a linear transformation takes spheres to ellipsoids, and is uniquely defined by how it does so.