That's exactly the question. I'm engineering undergrad so I didn't know the technical graph theory terms. It seems my construction is sightly better than the shown lower bound; is this significant, should I email this to the author? With some effort I'm convinced now you could show $k(n)=a(n)$.

Ah yes, something like that. The idea was to make the lines $f_1=k$ and $f_2=k$ parallel, by making both $f_1$ and $f_2$ a function only of the distance to two faces as you describe, or more simply a positive function of $x_m \cdot x$ (where $\cdot$ is the dot product). Indeed I think something like this gives a generalization, and a way to built $g$ such that $g(x)=(y,...,y) /iff x=x_m, x \cdot x_m = 0 or x=-x_m$. To that end, just choose two positive functions s.t. $f_1=h_1(x_m \dot x)$, $f_1=h_1(x_m \dot x)$, $h_1(y)=-h_1(y)$,$h_1(1)=-h(-1)=1=h_2(1)=-h_2(-1)$ that only intersect at 0,1,-1.

Just leaving as a bonus an interesting set I've found with $\mu(x) = 0$ en.wikipedia.org/wiki/Rep-tile#/media/… Thanks again for the fantastic proof!

Thanks for your answer. The need for linearly independent directions arises because otherwise we might not be able to pick a Lebesgue density point at the origin, is that right? This seems to settle the question for 3 simultaneous (non-trivial) symmetries! I'll have to look for approximate symmetries then.