Can each edge of a Wythoffian polytope be flipped by a reflection? A Wythoffian polytope $P\subset\Bbb R^d$ is an orbit polytope of a finite reflection group, that is,
$$P:=\mathrm{Orb}(\Gamma,x):=\mathrm{conv}\{Tx\mid T\in\Gamma\},$$
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).

Partial proof
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.
 A: Suppose $W$ is a fundamental domain of $\Gamma$, and $v \in W$ is a seed vertex.
By vertex transitivity, every edge of $P$ is $\gamma e$ for some edge $e$ incident to $v$ and some $\gamma \in \Gamma$, and if $e$ is bisected by a reflection $T \in \Gamma$,
then $\gamma T \gamma^{-1}$ is a reflection bisecting $\gamma e$.
In fact, every edge is the image of an edge leaving $v$ and passing through $W$, by the nature of the fundamental domain (possibly along part of the boundary of $W$.)
The other endpoint of $e$ cannot be in $W$, not even on its boundary,
because each point of $P$ is the image of exactly one point in the fundamental region. (Wikipedia cites Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Proposition 8.29 for this fact.)
At the point where the edge $e$ leaves $W$, it passes through a bounding reflection of $W$.
This reflection must carry $e$ onto some edge of $P$, and since it fixes one point in $e$, it must carry $e$ to itself.
A: Isn't that exactly the definition of being Wythoffian?
I.e. by virtue of kaleidoscopical construction the edges occur only as convex hulls of a seed point (vertex) and its mirror image wrt. to some mirror out of the reflection group. In fact one which occurs as boundary of the fundamental domain. Thence this very mirror fixes this very edge setwise, but reverses its orientation.
--- rk
