Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be radial. Then can $f$ necessarily be represented as $$ A(\|x\|)x = f(x), $$ for some function $z\mapsto A(z)$ from $\mathbb{R}$ into $SO(n)$?

Here, by radial I mean that there exists some $f_0\in C^1(\mathbb{R},\mathbb{R}^{n})$ such that $f(x)=f_0(\|x\|)$. So I'm curious if we can assume that $f_0$ is acting by matrix multiplication...

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    $\begingroup$ could you edit to define "radial", and make explicit any implicit continuity/smoothness assumption/requirement? $\endgroup$ – YCor Oct 5 at 16:53
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    $\begingroup$ As stated, no. Let's use $n=2$, and $f=\vec{e}_{1} =f_0 $, simply a constant unit vector in 1-direction. Then, for $x=\vec{e}_{1} $, $A$ is the identity map, whereas for $x=-\vec{e}_{1} $, $A$ is minus the identity map. However, $A$ must be the same in these two cases, contradiction. This seems too simple - perhaps there's still an error in the description? $\endgroup$ – Michael Engelhardt Oct 5 at 17:19
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    $\begingroup$ not clear: in the given definition, $f$ is radial iff its restriction on any sphere centered at the origin is constant, whereas $A(\|x\|)\cdot x$ is not. $\endgroup$ – Pietro Majer Oct 5 at 19:19
  • $\begingroup$ The given definition of "radial" immediately answers the question in the negative: for constant $\lVert x\rVert$, the action of a constant matrix on $x$ will not give a constant result. And what is meant by $f_0$ "acting"? (On what?) $\endgroup$ – gmvh Oct 6 at 13:52

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