TITLE: $SO(m+1)$-equivariant maps from $S^m$ to $S^m$
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: Proposition. Let $G$ be a group (acting on itself on the left), $X$ a $G$-set, and $f:G\to X$ a $G$-equivariant map. Then there exists a unique $x\in X$ such that $f(g)=gx$.

Proof: immediate.

Corollary: Let $G$ be a group, $H$ a subgroup, and $f$ a $G$-equivariant map $G/H\to G/H$. Then $f$ is a permutation of $G/H$, and has the form $f(gH)=gqH$, where $q\in N_G(H)$, and $q$ is well-determined as element of $N_G(H)/H$.

Proof: First compose as a map $G\to G/H$: then we deduce from the proposition that $f(gH)=gqH$ for some $q\in G$ and all $g\in G$; we see that $q$ is determined modulo right multiplication by $H$. Since $f(gH)=f(ghH)$ for all $h\in H$, we deduce $gqH=ghqH$ for all $h\in H$ and $g\in G$, that is, $q^{-1}hq\in H$ for all $h\in H$. This precisely means $q\in N_G(H)$.

Corollary: in the setting of the question, we only have $\pm$ identity.

Proof: write $S^m=\mathrm{SO}(m+1)/\mathrm{SO}(m)$. By the corollary, we need to determine the normalizer of $\mathrm{SO}(m)$. Since $m\ge 2$, the latter preserves a unique line, hence this line is preserved by this normalizer, which equals $\mathrm{S}(\mathrm{O}(m)\times\mathrm{O}(1))$ and contains $\mathrm{SO}(m)$. This corresponds to the two desired equivariant maps.
Such a reasoning extends to various other homogeneous spaces.