Symmetries and faces of the associahedron 
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*The dihedral group of order $2n+2$ acts on $K_n$, the ($n-2$)-dimensional associahedron.  Are there any other symmetries?  References?


*Does the answer to 1 change if we restrict to just the 1-skeleton of $K_n$?  References?


*It is "obvious" that any simple circuit (simple closed walk, simple closed path, whatever terminology you prefer) of length 4 or 5 is a 2-dimensional face of $K_n$.  Is this true?  Proof?  Reference?
 A: I have just seen this question, while looking for something else. The answer to 1 is indeed "no", and an explicit proof appears in Lemma 2.2 of Ceballos, Santos, and Ziegler - Many non-equivalent realizations of the associahedron (not surprisingly, it follows the same ideas as Patricia's).
A: The answer to question 1 is no.  A reference for this is:
Carl Lee, The associahedron and triangulations of the $n$-gon, European Journal of Combinatorics, 10 (1989), no. 6, 551--560.
The answer to question 3 is yes.  I think this is clear from the viewpoint where you think of vertices of the associahedron as triangulations of an $(n+1)$-gon and you obtain higher dimensional faces containing such a vertex by deleting edges from the triangulation.  This is the viewpoint e.g. discussed by Carl Lee.  A 4-cycle involving a vertex $v$ of the associahedron implies that the two edges $e_1,e_2$ in the 4-cycle containing $v$ correspond to the deletion of a pair of edges $E_1,E_2$ from the triangulation corresponding to $v$ such that the concurrent deletion of $E_1, E_2$ yields two quadrilateral regions in the resulting subdivision; a 5-cycle involving a vertex $v$ of the associahedron likewise results from two edges $E_1, E_2$ of the corresponding triangulation whose concurrent deletion yields a single pentagonal region.  In either case, the 4-cycle or 5-cycle then clearly bounds a face of the associahedron, namely the one given by the subdivision in which $E_1$ and $E_2$ are deleted from the triangulation corresponding to $v$.  
${\bf Edit:}$  I just realized we can deduce that the answer to 2 is also no, by virtue of a result of Gil Kalai.  Kalai proved that any $d$-dimensional simple polytope is determined by its 1-skeleton.  So we can use that the associahedron is a simple polytope to see that its 1-skeleton can't have any extra symmetries not present in the associahedron itself.
