My Shibuya repository now contains unit-distance embeddings in the plane of all cubic symmetric graphs to $120$ vertices, except the first two ($K_4$ and $K_{3,3}$) which do not have this property, as well as some not-so-symmetric cubic graphs (e.g. the McGee graph from this question and the Gray graph) and a non-degenerate Holt graph.
While I was making these embeddings (most of them via a semi-automatic program in Shibuya) I had a 2018 paper by Frankl, Kupavskii and Swanepoel in mind, which says that any connected graph with maximum degree $d$ can be unit-distance embedded in $\mathbb R^d$ – except $K_{3,3}$.
$K_4$ is obviously unit-distance embeddable in $\mathbb R^3$ with the maximum possible symmetry of $S_4$ – all automorphisms of the graph are symmetries of the embedding. That made me wonder if $K_{3,3}$ could be unit-distance embedded in $\mathbb R^4$ with its maximum symmetry of $(S_3×S_3)\rtimes C_2$. It is not too hard to find one (each vertex on the left is connected to all on the right), and indeed generalising this construction shows that all graphs with chromatic number $k$ may be embedded into $\mathbb R^{2k}$: $$\frac1{\sqrt2}(1,0,0,0)\qquad\frac1{\sqrt2}(0,0,1,0)\\ \frac1{\sqrt2}(-1/2,\sqrt3/2,0,0)\qquad\frac1{\sqrt2}(0,0,-1/2,\sqrt3/2)\\ \frac1{\sqrt2}(-1/2,-\sqrt3/2,0,0)\qquad\frac1{\sqrt2}(0,0,-1/2,-\sqrt3/2)$$ This embedding's symmetries are generated by the following orthogonal matrices: $$\begin{bmatrix} 1&0&0&0\\ 0&-1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}\qquad\begin{bmatrix} -1/2&\sqrt3/2&0&0\\ \sqrt3/2&1/2&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}\qquad\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&-1 \end{bmatrix}\qquad\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&-1/2&\sqrt3/2\\ 0&0&\sqrt3/2&1/2 \end{bmatrix}\qquad\begin{bmatrix} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0 \end{bmatrix}$$ But having read about Coxeter notation for point groups in arbitrary dimensions, I'm having trouble representing the above symmetries as a subgroup of a Coxeter group. Without the last matrix this would be $[3]\times[3]$ or $[3,2,3]$, but this last matrix has determinant $+1$ and I can't "fold" it into the other four matrices to make a rank-$4$ group of only reflections.
How can the symmetry group generated by all five of the above matrices be written in Coxeter notation in the shortest possible way?
Conway and Smith's On Quaternions and Octonions contains a listing of all four-dimensional point groups, treating vectors in $\mathbb R^4$ as quaternions. For odd $n$ consider the following embedding of $K_{n,n}$, which when $n=3$ is equivalent to the embedding of $K_{3,3}$ in question after an axis swap and a scaling. Each point obtained from $f=i^n$ ("ring A", blue in the picture below) is connected to every point obtained from $f=j$ ("ring B", orange): $$\{e^{2im\pi/n}f:0\le m<n,f\in\{i^n,j\}\}$$
Then the embedding's symmetry group has no Coxeter notation but is denoted in Conway & Smith as $$+\frac14[D_{4n}×\overline D_{4n}]\cdot2_1$$ It is generated by the following $5$ quaternion operators: $$[-e_n,1],[-1,e_n],[e_{2n},j],[j,e_{2n}],*[-1,1]$$ $$e_n=e^{i\pi/n},[l,r]:q\to\overline lqr,*[l,r]:q\to\overline l\overline qr$$ For $n=2p$ the embedding of $K_{n,n}$ is simpler in description: $$\{e^{im\pi/p}f:0\le m<n,f\in\{1,j\}\}$$ Its symmetry group has Conway & Smith notation $$\pm\frac12[\overline D_{4p}×\overline D_{4p}]\cdot2$$ with $11$ generators as follows: $$[\pm e_p,1],[\pm j,1],[\pm1,e_p],[\pm1,j],[\pm e_n,e_n],*[1,1]$$