The dihedral group of order $2n+2$ acts on $K_n$, the $n2$dimensional associahedron. Are there any other symmetries? References?
Does the answer to 1 change if we restrict to just the 1skeleton 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 2dimensional face of $K_n$. Is this true? Proof? Reference?



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, 551560. 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 4cycle involving a vertex $v$ of the associahedron implies that the two edges $e_1,e_2$ in the 4cycle 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 5cycle 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 4cycle or 5cycle 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 1skeleton. So we can use that the associahedron is a simple polytope to see that its 1skeleton can't have any extra symmetries not present in the associahedron itself. 


I have just seen this question, while looking for something else. The answer to 1 is indeed "no", and an explicit proof appears in http://arxiv.org/abs/1109.5544 (no surprisingly, it follows the same ideas as Patricia's) 

