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Suppose that there are two combinatorially equivalent (convex) polytopes $P_1,P_2\subset\Bbb R^d$, that is, both with the same face lattice $\mathcal L$. The symmetry group $\mathrm{Aut}(P_i)\subset\mathrm O(\smash{\Bbb R^d})$ of $P_i$ induces a group $\Sigma_i\subseteq\mathrm{Aut}(\mathcal L)$ of combinatorial symmetries on the face lattice.

Now, we can generate the larger group $\Sigma:=\langle \Sigma_1,\Sigma_2\rangle\subseteq\mathrm{Aut}(\mathcal L)$ of symmetries of $\mathcal L$.

Question: Is there also a realization of $\mathcal L$ as a polytope $P\subset\smash{\Bbb R^d}$ whose symmetry group $\mathrm{Aut}(P)$ induces (at least) $\Sigma$?

As an example, here are two quadrangles, one is vertex-transitive, one is edge-transitive, and they combine to a quadrangle that is vertex- and edge-transitive.

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  • $\begingroup$ Your previous question seems very related: mathoverflow.net/questions/311877/… $\endgroup$ Commented Jun 9, 2020 at 21:24
  • $\begingroup$ @SamHopkins Please help me if I am missing something, but the new thing in above questios is, that I already know that the symmetries I am asking for can in fact be realized. $\endgroup$
    – M. Winter
    Commented Jun 9, 2020 at 21:25
  • $\begingroup$ Sure, I'm just saying, you're again asking for a bigger set of symmetries. $\endgroup$ Commented Jun 9, 2020 at 21:28
  • $\begingroup$ @M.Winter Do you require equality to hold, i.e. must Aut(P) be equal to $\Sigma$, or merely contain $\Sigma$? $\endgroup$ Commented Jun 9, 2020 at 22:48
  • $\begingroup$ @Adam Indeed, I only require that $\Sigma$ is contained. Thanks for the note. $\endgroup$
    – M. Winter
    Commented Jun 9, 2020 at 22:56

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