A Wythoffian polytope $P\subset\Bbb R^d$ is an orbit polytope of a finite reflection group, that is,
where $\Gamma$ is a finite reflection group, and $x\in\Bbb R^d$.
Question: Is it true, that if $e$ is an edge of $P$, then there is a reflection $T\in\Gamma$ that fixes $e$ set-wise but flips its orientation?
I am pretty sure that this is true, but I cannot find a concise proof from first principles. I would be most happy with a short self-contained argument (possibly using some well-known properties of reflection groups).
Suppose that the generator $x\in\Bbb R^d$ was chosen from the interior of a Weyl chamber of $\Gamma$ (the resulting polytope is sometimes called a $\Gamma$-permutahedron or omnitruncated uniform polytope). Since $\Gamma$ acts regularly on the Weyl chambers, every Weyl chamber contains a single vertex. In particular, if $e$ is an edge of $P$, its end vertices are in different chambers and the edge must cross a bounding reflection hyperplane of the chamber. Since reflection on this hyperplane is a symmetry of $P$, the edge must be perpendicular to the hyperplane and the reflcetion on it must flip its orientation.
My hope is that this can be extended to an argument for general placements of $x$. Each Wythoffian polytope can be obtained as a "limit" (in some sense) of such $\Gamma$-permutahedra, and maybe this sufficec to prove the statement. However, I was not able to make a clear case for why the edges of the resulting polytope are "limits" of edges of the $\Gamma$-permutahedra.