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.
