Let $G=SO(m+1)$ , $m \geq 2$, act in the standard way on $S^m$.


Let $F:S^m \to S^m$ be a $G$-equivariant map, i.e., $g F(g^{-1}x) =F(x)$ for all $x \in S^m$ and $g \in G$.



> 
> **Question 1:** Is F the identity map?
> 
> 
> 


If the answer is negative: Is $F$ an isometry?



$SO(m+1)$-equivariant maps from $S^m$ to $S^m$

> 
> **Theorem.** $F:\mathbb{S}^m\to \mathbb{S}^m$, $m\geq 2$, is $SO(m+1)$ equivariant if and only if $F=\operatorname{Id}$ or $F=-\operatorname{Id}$.
> 
> 
> 


Let me write a **very detailed proof** that only requires a basic knowledge of linear algebra.


**Proof.** 
It is easy to see that both $F=\operatorname{Id}$ and $F=-\operatorname{Id}$
are $SO(m+1)$ equivariant so it remains to prove that if $F$ is equivariant, then 
$F=\operatorname{Id}$ or $F=-\operatorname{Id}$.


Let $e\_1,e\_2,\ldots, e\_{m+1}$ be the standard orthogonal basis of $\mathbb{R}^{m+1}$. If
$[\rho\_{jk}]$ is the matrix representation of $\rho\in SO(m+1)$, then the condition
$$
F(\rho (x))=\rho (F(x))
$$
reads as
$$
(\*)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ F\_j(\rho(x))=\sum\_{k=1}^{m+1}\rho\_{jk}F\_k(x),
\quad
j=1,2,\ldots,n,
$$
where $F(x)=(F\_1(x),\ldots,F\_n(x))$.


Let $F\_1(e\_1)=c$. Consider all $\rho\in SO(m+1)$ such that $\rho(e\_1)=e\_1$.
This condition means that the first column of the matrix $[\rho\_{jk}]$ equals $e\_1$, i.e.
$\rho\_{11}=1$, $\rho\_{j1}=0$, for $j>1$. Since columns are orthogonal, for $k>1$ we have
$$
0=\sum\_{j=1}^{m+1}\rho\_{j1}\rho\_{jk}=\rho\_{1k}\, .
$$
Thus
$$
\rho =
\left[
\begin{array}{cccc}
1 & 0 & \ldots & 0 \\
0 & \rho\_{22} & \ldots & \rho\_{2,m+1} \\
\vdots & \vdots & \ddots & \vdots \\
0 & \rho\_{m+1,2} & \ldots & \rho\_{m+1,m+1}
\end{array}
\right]\, ,
$$
where $[\rho\_{jk}]\_{j,k=2}^{m+1}$ is the matrix of an arbitrary transformation in
$SO(m)$ (rotation in the
$m$-dimensional subspace orthogonal to $e\_1$).


For $x=e\_1=\rho(e\_1)=\rho(x)$ and $j\geq 2$ identity ($\*$) yields
$$
F\_j(e\_1)=\sum\_{k=1}^{m+1} \rho\_{jk} F\_k(e\_1) =
\sum\_{k=2}^{m+1} \rho\_{jk}F\_k(e\_1)\, ,
$$
and hence
$$
\left[
\begin{array}{c}
F\_2(e\_1) \\
\vdots \\
F\_{m+1}(e\_1)
\end{array}
\right]
=
\left[
\begin{array}{ccc}
\rho\_{22} & \ldots & \rho\_{2,m+1} \\
\vdots & \ddots & \vdots \\
\rho\_{m+1,2} & \ldots & \rho\_{m+1,m+1}
\end{array}
\right]\,
\left[
\begin{array}{c}
F\_2(e\_1) \\
\vdots \\
F\_{m+1}(e\_1)
\end{array}
\right]\, .
$$
That means the vector $[F\_2(e\_1),\ldots,F\_{m+1}(e\_1)]^T$
is fixed under any transformation $SO(m)$ of
$\mathbb{R}^{m}$, so it must be a zero vector, i.e.
$$
F\_2(e\_1)=\ldots=F\_{m+1}(e\_1)=0\, 
$$
so 
$$
F(e\_1)=(c,0,\ldots,0),
\quad
c=\pm 1.
$$
Now formula ($\*$) for any $\rho\in SO(m+1)$ and
$x=e\_1$, takes the form
$$
F\_j(\rho(e\_1))=\rho\_{j1}F\_1(e\_1)=\pm\rho\_{j1}\, .
$$
Let $x\in \mathbb{S}^m$ and let
$\rho\in SO(m+1)$ be such that $\rho(e\_1)=x$. Then
$\rho\_{j1}=x\_j$, $j=1,2,\ldots,n$ and hence
$$
F\_j(x)=\pm\rho\_{j1}=\pm x\_j,
\quad
F(x)=\pm x.
$$
Therefore $F(x)=x$ or $F(x)=-x$.


