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Yes, all functions in $X_a$ are still symmetric wrto the line generated by $a$. And all traces of these functions on the affine hyperplanes orthogonal to $a$ are radially symmetric in dimension $N-1$.

edit. Consider any pair $x$ and $x'$ in $\mathbb{R}^N$ with $|x|=|x'|=1$ and $a\cdot x=a\cdot x'$. Then $|x-\lambda a |=|x'-\lambda a|\ ,$ so $u(x)=u(x')$ for all $\lambda$ and all $u\in X_{\lambda,a}$. And this symmetry is preserved taking the linear span and the closure: $u=u\circ R$ holds for any $u\in X_a$ any for any orthogonal $R$ that fixes $a\ .$
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Yes, all functions in $X_a$ are still symmetric wrto the line generated by $a$. All And all traces of these functions on the affine hyperplanes orthogonal to $a$ are radially symmetric in dimension $N-1$.

edit. Consider any pair $x$ and $x'$ in $\mathbb{R}^N$ with $|x|=|x'|=1$ and $a\cdot x=a\cdot x'$. Then $|x-\lambda a |=|x'-\lambda a|\ ,$ so $u(x)=u(x')$ for all $\lambda$ and all $u\in X_{\lambda,a}$. And this symmetry is preserved taking the linear span and the closure: $u=u\circ R$ holds for any $u\in X_a$ any any orthogonal $R$ that fixes $a\ .$
show/hide this revision's text 2 added 283 characters in body

Yes, all functions in $X_a$ are still symmetric wrto the line generated by $a$. All traces of these functions on the affine hyperplanes orthogonal to $a$ are radially symmetric in dimension $N-1$.

edit. Consider any pair $x$ and $x'$ in $\mathbb{R}^N$ with $|x|=|x'|=1$ and $a\cdot x=a\cdot x'$. Then $|x-\lambda a |=|x'-\lambda a|\ ,$ so $u(x)=u(x')$ for all $\lambda$ and all $u\in X_{\lambda,a}$. And this symmetry is preserved taking the linear span and the closure.
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