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If you arrange the edge skeletons of a convex regular pentagon and a regular pentagram in the right way and connect each vertex of the former to that of the latter lying under it, the result (the Petersen graph) is significantly more symmetric than you might expect, and in fact is a symmetric graph. Does the same thing happen if you use a convex regular icosahedron (or great dodecahedron) and a small stellated dodecahedron (or great icosahedron) instead of the pentagons?

If so, the full automorphism group should be of order 2880 (Edit: or 1440).

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    $\begingroup$ I think all the polyhedron you suggested have 12 vertices, but there is no transitive group of order 2880 and degree 12. (But there are some such groups of degree 20.) $\endgroup$
    – verret
    Commented Feb 2, 2023 at 18:14
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    $\begingroup$ @verret I understood the suggested graph to have 24 vertices, just as the Petersen graph has 10 vertices, not 5. $\endgroup$
    – Andreas Blass
    Commented Feb 2, 2023 at 18:23
  • $\begingroup$ Thanks for the correction. Here's another attempt: according to math.auckland.ac.nz/~conder/AllSmallETgraphs-upto47-summary.txt, there is no edge-transitive graph of order $24$ with automorphism group $2880$. $\endgroup$
    – verret
    Commented Feb 4, 2023 at 0:37
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    $\begingroup$ @Daniel Sebald : there is no edge-transitive graph of order $24$ with automorphism group of order $1440$ either. There are only $65$ connected edge-transitive graphs of order $24$, so maybe you could check them to see if anything looks like what you are after? $\endgroup$
    – verret
    Commented Feb 4, 2023 at 7:15

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Rather than answering the question you literally asked (since verret in the comments showed the answer is "no"), I would suggest reconsidering the construction you want to generalize. Instead of viewing the Petersen graph as the outcome of connecting pentagon graphs in a certain way, it is more natural to view the Petersen graph as the quotient of the edge graph of a dodecahedron under the identification of antipodes. This construction doesn't privilege any pentagons, and the $S_5$ symmetry is mostly inherited from the symmetry of the dodecahedron. If you want a more natural generalization to a higher dimension, perhaps consider quotients of highly symmetric polytopes under subgroups of their symmetries.

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