$SO(m+1)$-equivariant maps from $S^m$ to $S^m$ 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?
 A: 
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 $F(x)=-x$.
A: 
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.
A: $G$ acts transitively on $S^m$, so if you know the image of the north pole $N = (0,\dotsc,0,1)$, there is for every other point $p\in S^m$ an element $g\in S^m$ such that $p = g\cdot (0,\dotsc,0,1)$. Thus you will have $F(p) = F\bigl(g\cdot (0,\dotsc,0,1)\bigr) = g\cdot F(0,\dotsc,0,1)$, i.e., knowing the image under $F$ of one single point already determines the map $F$.
Now for the question whether $F$ is the identity.  $N$ is fixed by the subgroup $H = SO(m)$, thus it follows that $F(N)$ also needs to be fixed by the same subgroup, but the only elements in $S^{m+1}$ fixed by $H$ are the north and the south pole.
Now if you take into account that $g\cdot (-p) = - g\cdot p$, it follows should follow that $F$ is either the identity or the antipodal map.
