Isometric action on $S^n$ Let $S^n$ be the n dimensional sphere. For $n=2k+1$ odd, we identify $S^n$ as subset of $\mathbb{C}^{k+1}$.
Furthermore we can define the action 
$$\Psi: S^1 \times S^n \to S^n, (c,(z_0, \dots, z_n)) \mapsto (cz_0, \dots, cz_n).$$
If we write $z_j$ in polarcoordinates, then $z_j = r_je^{i\phi_j}$ and $c=e^{i\alpha}$. Therefore the action is given by $\Psi_c(r_0e^{i\phi_0}, \dots, r_n e^{i\phi_n})=(r_0e^{i(\phi_0+\alpha)}, \dots, r_ne^{i(\phi_n+\alpha)})$.
I read everwhere that this action is isometric, w.r.t. the canonical metric on $S^n$, but I'm not sure how to prove it using a local representation of the metric.
 A: (1) If $R$ is $\theta$-rotation on ${\bf R}^2$, then consider $$\overline{R} := \underbrace{(R,\cdots, R)}_{k+1-times} : {\bf R}^{2n+2}={\bf C}^{n+1}\rightarrow {\bf R}^{2n+2} $$ That is $\overline{R}\in SO(2n+2)$ is an isometry on $({\bf R}^{2n+2},g)$ where $g$ is a canonical metric
(2) Clearly $\overline{R}$ acts on $M:=S^{2n+1}$ For any two points $p,\ q\in M$, they have distance $a=d(p,q)$ where $d$ is an induced distance from $g$ and $a=\angle (p,q)$ 
Note that isometry on ${\bf R}^{2n+2}$ preserves angle so that $$ d(\overline{R}p,\overline{R}q)=\angle ( \overline{R}p,\overline{R}q )=d(p,q) $$
Hence it is an isometry on $S^{2n+1}$
[Add] I will add a proof of Thomas's comment :
$G=S^1$ acts on $M={\bf R}^{2n+2}$ isometrically And $G$ has an
invariant submanifold $N=S^{2n+1} \subset M$ Then $G$-action on $N$
is isometric :
Proof : $N$ has a distance function $d$ : For $p,\ q\in N$, $d(p,q)=
\inf_c {\rm length}\ c$ where $c$ is a curve from $p$ to $q$ in $N$.
For $g\in G$, note that $$ | \frac{d}{dt} g(c(t)) | = |\frac{d}{dt}
c(t)| \Rightarrow {\rm length}\ g(c) ={\rm length}\ c $$
So $$ d(g(p),g(q))= d(p,q) $$
