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Let $f:S^n \to S^n$ be a homeomorphism. I know the result that a rigid motion in $\mathbb R^{n+1}$ is always linear, but can we get more information from the assumption that $f:S^n \to S^n$ is a homeomorphism?

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    If you have a linear function on $\Bbb R^{n+1}$ that you know restricts to a homeomorphism on the unit $n$-sphere, then you immediately know that the linear map can only be a combination of reflections and rotations. Is that what you're after? – 2017-02-16
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    @Arthur No, I only know that there is such a homeomorphism and ask if it can be extended to a linear map on $\mathbb R^{n+1}$. – 2017-02-16
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    @TiWen I don't undertstand the question. If the homeomorphism is not a restriction of a linear map (most homeomorphism of the sphere are not linear), how could you extend it to a linear map? – 2017-02-16
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    @PeterFranek What is an example of nonlinear homeomorphism of sphere? – 2017-02-16
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    @TiWen For example, $f(x)=\frac{x+(0.1 \sin(x_1),0,0,\ldots,0)}{\| x+ (0.1\sin(x_1),0,0,\ldots, 0)\|}$ where $x=(x_1,\ldots x_n)$. – 2017-02-16

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The question is hard to understand in its current form, but a few remarks:

  1. Not every homeomorphism of the sphere is linear.
  2. Every homeomorphism of the sphere extends to a homeomorphism of $\Bbb R^{n+1}$ (see Tsemo Aristide answer)
  3. If you insisted on smoothness and diffeomorphisms extension, the problem is far more complicated and depends very much on the dimension $n$.
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$\newcommand{\C}{\mathbb{C}}$$\renewcommand{\theta}{\vartheta}$$\newcommand{\R}{\mathbb{R}}$Consider $S^{1} \subseteq \R^{2}$. The map $$ f(e^{i \theta}) = e^{i \theta^{2}/2 \pi} $$ is a homeomorphism of $S^{1}$, as $\theta \mapsto \dfrac{\theta^{2}}{2 \pi}$ is a homeomorphism on the interval $[0, 2 \pi]$.

However, it maps $(1, 0) = e^{i 0}$ to itself, but its multiple $(-1, 0) = e^{i \pi}$ to $e^{i \pi/2} = (0, 1)$.

Here I have identified $\R^{2}$ with $\C$.

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Let $f:S^n\rightarrow S^n$ an homeomorphism. define $F:R^n\rightarrow R^n$ by $F(x)=\|x\|f({x\over{\|x\|}}), x\neq 0$, $F(0)=0$.

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    Why $F(x+y)=F(x)+F(y)$? – 2017-02-16