Let $d\in\mathbb N$, $\Omega\subseteq\mathbb R^d$ and $f:\Omega\to\mathbb R$.
Remember that $f$ is called radial if $f(x)=f(y)$ for all $x,y\in\Omega$ with $\|x\|=\|y\|$. Let $O(d)$ denote the orthogonal group of dimension $d$.
How can we show that $f$ is radial iff $f=f\circ\left.U\right|_\Omega$ for all $U\in O(d)$?
The first implication is easy to show: If $f$ is radial and $U\in O(d)$, then $f(x)=(f\circ U)(x)$ for all $x\in\Omega$, since $\|Ux\|=\|x\|$ for all $x\in\mathbb R^d$.
However, I don't know how we can establish the other implication. So, how can we do it?
(I've only seen the claim in the special case $\Omega=B_r(0)$ for some $r>0$. I don't think that this particular setting is important, but since I'm failing to show the other implication, feel free to restrict to this case, if I'm wrong and this is necessary.)