Isometries of $\mathbb{S}^n$ How to prove so elementary (elementary = without using the concept of geodesic) that an isometry of $\mathbb{S}^n$ is a restriction on $\mathbb{S}^n$   of  an isometry of $\mathbb{R}^{n+1}$  ?
EDIT:
You will isometry  with respect metric is induced by $\mathbb{R}^{n+1}.$
 A: 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.
A: 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.
A: A correct presentation would be pretty long. Let us start with the positive orthant, all $x_j \geq 0$ in $ \mathbb S^n.$ The fact that the unit sphere is given by by a sum of squares equalling one amounts to direction cosines. That is, given the standard orthonormal basis $e_j,$ we are saying that
$$ \sum \cos^2 \theta_j = 1,  $$
where $\theta_j$ is the angle between our favorite point $x$ on the sphere and $e_j.$ This is not a surprise, as
$$  \cos \theta_j = x \cdot e_j = x_j,  $$ the $j$th coordinate, and we are just saying the sum of the squares of the coordinates is $1.$ 
Next, we need a definition of distance on the metric space  $ \mathbb S^n.$ I say
$$ \mbox{AXIOM} \; \; \; d(x,y) = \arccos (x \cdot y).  $$
Another way of saying $\sum x_j^2 = 1$ is then
$$ 1 =  \sum (x \cdot e_j)^2 = \sum \cos^2 \theta_j = \sum \cos^2 d(x,e_j).$$
Here is the part that would take forever to write out:
AXIOM: If I take a finite set
$$ \{\epsilon_j\}, \; \; \epsilon_j \geq 0, \; \; \sum \epsilon_j = 1,    $$
there is exactly one point $x$ in the positive orthant such that each
$$  (x \cdot e_j)^2 =  \epsilon_j.    $$ 
Sketch: the isometry maps the positive orthant to something....
5:02 pm Pacific time. I will probably pick this up later...
