4 added 2 characters in body

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:}

${\bf Edit:}$ 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.

3 added 362 characters in body

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.

2 changed $n$-gon to $(n+1)$-gon

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 $n$-gon (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\$. I'm sorry that's a mouthful.

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