Coxeter notation for the symmetries of the maximally symmetric unit-distance embedding of $K_{3,3}$ in $\mathbb R^4$ 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?

 A: 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:
$$[\pm e_p,1],[\pm j,1],[\pm1,e_p],[\pm1,j],[\pm e_n,e_n],*[1,1]$$
