**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 $F(x)=-x$.