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Let $n\in\mathbb{N}$, a mapping $T:\mathbb{R}^n\to \mathbb{R}^n$ is called conformal transform if it preserves the angle between curves. Namely, if $u,v :I\to \mathbb{R}^n$ such that $u(0)= v(0)$. Where $I\subset \mathbb{R}$ is an open interval containing $0$. then \begin{equation} \cos^{-1} \left(\frac{\overset{.}{u}(0)}{\|\overset{.}{u}(0)\|}\cdot \frac{\overset{.}{v}(0)}{\|\overset{.}{v}(0)\|}\right) =\cos^{-1} \left(\frac{\overset{.}{Tu}(0)}{\|\overset{.}{Tu}(0)\|}\cdot \frac{\overset{.}{Tv}(0)}{\|\overset{.}{Tv}(0)\|}\right) \end{equation} where $x\cdot y $ is the usual scalar product of $\mathbb{R}^n$ and $\|.\|$ is the corresponding Euclidean norm.

Question It is stated as obvious result in the book of Lieb Loss and Eliot (Analysis second edition) page 120 Lemma 4.8 that any such Conformal Transformation is product of Euclidean group (Please what is the Euclidean group?), scaling map, translation. In short it can be written as product of similitudes.

As well the result is well known in the complex plane thank to the Mobius maps.

I would like to know where or how can one find or prove of the above statement in higher dimension $n\ge3$?

Any prove or help is welcome.

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    I think you meant $\cos^{-1}$ to turn scalar products of normalized vectors into angles. But that part is actually irrelevant; you might as well require the dot products themselves to be equal. The Euclidean group would be the group of all isometric (i.e. length-preserving) transformations. I'm surprised they list translations separately, as I'd include them. – 2017-02-07
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    You are right I just fixed it . thanks – 2017-02-07
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    You are right I just fixed it . As well we can get rid of $\cos^{-1}$. I think also one can include translation as well in the Euclidean group. but they stated it that way – 2017-02-07

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This is hard to prove directly, but is an easy corollary of Liouville's theorem. Once you have this theorem, use the general formula for the Moebius transformation given in the link: $$ f(x)= b + \frac{\alpha A(x-a)}{|x-a|^{\epsilon}}, \epsilon\in \{0, 2\}, $$ where $A$ is an orthogonal matrix, $\alpha$ is a scalar and $a, b$ are vectors in $R^n$. Since in your case $f$ sends $R^n$ to itself, $\epsilon=0$ and you are done.