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In Are Your Polyhedra the Same as my Polyhedra? by Branko Grünbaum, the author (sort of) defines the symmetries of a polygon by saying

The historically and practically most important classes are defined by symmetries, that is, by isometric transformations of the plane of the polygon that map the polygon onto itself. In case some of the vertices coincide, symmetries should be considered as consisting of an isometry paired with a permutation of the vertices.

In this paper, polygons are defined as cyclic sequences of points without any further constraints (so there can be "edges" going from a point to itself, self-intersections, and so forth).

The example given is a "quadrangle" consisting of an equilateral triangle where one vertex is regarded as two separate, consecutive vertices (Figure 2 on p. 465). The caption reads

This polygon looks like an equilateral triangle, but is in fact a quadrangle with two coinciding vertices. Besides the identity, the only symmetry it admits is a reflection (in a vertical mirror through the coinciding vertices 3 and 4) paired with the permutation $(1 2)(3 4)$.

I don't quite understand what the actual definition he's hinting at is. If I had to guess it would be something like this:

A symmetry of a polygon $V_1, \ldots, V_n$ in a space $X$ is an isometry $f : X \to X$ together with a permutation $\sigma \in \mathfrak{S}_n$ such that $f(V_i) = V_{\sigma(i)}$.

But then we seem to have three symmetries other than the identity here: if $f$ is the reflection over the $Y$-axis, then we may as well take $\sigma = (1 2)$ as $\sigma = (1 2)(3 4)$. And then for that matter we can throw in their product, consisting of the identity paired with the permutation $(3 4)$.

Does Grünbaum actually mean to say that a symmetry is an isometry for which such a permutation exists, rather than an isometry together with such a permutation?

Or is there some implicit condition that rules out the two "symmetries" I've described? If we think about this "quadrangle" as the degenerate limit of a sequence of isosceles trapezoids where one side is gradually contracted away then the nondegenerate trapezoids only have the one symmetry. Earlier, Grünbaum stipulates that, for a satisfactory definition of a "polyhedron,"

[t]he combinatorial type should remain constant under continuous changes of the polyhedron

without ever saying exactly what "combinatorial type" means.

Does anyone know exactly what the definition is supposed to be?

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Actually, I think I found the answer. Later in the paper I found a corresponding definition for the three-dimensional case:

A symmetry of a (geometric) polyhedron is a pairing of an isometric mapping of the polyhedron onto itself with an automorphism of the underlying abstract polyhedron.

And a few pages earlier,

A symmetry of an abstract polyhedron is an automorphism induced by incidence-preserving permutations of the vertices, the edges, and the faces.

I'm not clear on what "induced by" means here, but if we consider the underlying abstract graph of the "quadrangle" (which is just a 4-cycle), then $(1 2)(3 4)$ is a graph automorphism but $(1 2)$ and $(3 4)$ are not.

So I suspect the definition is meant to be:

A symmetry of a polygon $\{V_1, \ldots, V_n\}$ in a space $X$ is an isometry $f : X \to X$ together with a dihedral permutation $\sigma \in D_{2n}$ such that $f(V_i) = V_{\sigma(i)}$.

Does that sound right to everyone else?

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