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Theorem. $F:S^m\to S^m$, $m\geq 2$, 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 $x\in S^m$ 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 $F(x)=-x$.

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 $F(x)=-x$.

Theorem. Under the above assumptions $F(x)=x$ or $F(x)=-x$.

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

Proof. 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 $x\in S^m$ 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 $F(x)=-x$.

Theorem. $F:S^m\to S^m$, $m\geq 2$, 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 $x\in S^m$ 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 $F(x)=-x$.

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 an arbitrary orthogonal transformation of $\mathbb{R}^{m}$, so it must be a zero vector, i.e. $$ F_2(e_1)=\ldots=F_{m+1}(e_1)=0\, . $$ Now formula (*) for any $\rho\in SO(m+1)$ and $x=e_1$, takes the form $$ F_j(\rho(e_1))=\rho_{j1}F_1(e_1)=c\rho_{j1}\, . $$ Let $x\in S^m$ 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 $$ F_j(x)=c\rho_{j1}=cx_j, \quad F(x)=cx. $$ Therefore $c=\pm 1$ and we have that $F(x)=x$ of $F(x)=-x$.

Theorem. Under the above assumptions $F(x)=x$ or $F(x)=-x$.

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

Proof. 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 $x\in S^m$ 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 $F(x)=-x$.