Let $N \geq 3$. Does there exist a set in $\mathbb{R}^N$ which is 8-fold symmetric with respect to any coordinate plane $(x_i,x_j)$, and which is not radially symmetric (i.e., not a sphere, ball, spherical shell, or their union, all centred at the origin)?
It seems that in the three-dimensional case, radially symmetric objects are the only examples of 8-fold symmetric sets wrt any coordinate plane. Maybe there is some general way how to establish this result rigorously in all dimensions?
This question is related to and motivated by my previous question in which the 8-fold symmetry assumption wrt some coordinate plane is imposed.
No!
Claim: There exists a nonempty countable subset of $\mathbb{R}^N-\{0\}$ which is invariant under all $\pi/4$ rotations on any coordinate 2-planes.
Proof: There are only finitely many coordinate 2-planes, so the group generated by $\pi/4$-rotations on them fixing the other $N-2$ directions is countable. Now just pick any point and look at its orbit. QED.
This clearly is not radially symmetric, since $SO(N)$ is transitive on the sphere $S^{N-1}(r)$ of radius $r$.
As a concrete example, consider $S=(\mathbb{Q}[\sqrt{2}])^N\subset\mathbb{R}^N$. A $\pi/4$-rotation in $(x_{n+1},x_{n+n'+2})$-plane has matrix representation $$ \begin{pmatrix} I_{n}\\ &\frac1{\sqrt2}&&\pm\frac1{\sqrt2}\\ &&I_{n'}\\ &\mp\frac1{\sqrt2}&&\frac1{\sqrt2}\\ &&&&I_{N-n'-n-2} \end{pmatrix} $$ so is an element of $SL_N(\mathbb{Q}[\sqrt2])$ hence must preserve $S$.