$[n]$ is the set $\{0,1\}^n$ equipped with the product order $(\epsilon\_1,\dots,\epsilon\_n) \leq (\eta\_1,\dots,\eta\_n)$ if and only if $\forall i=1,\dots,n$, $\epsilon\_i \leq \eta\_i$. Let $$d((\epsilon\_1,\dots,\epsilon\_n),(\eta\_1,\dots,\eta\_n)) = \sum\_{i=1}^{i=n}|\epsilon\_i-\eta\_i|.$$ A set map $f$ from $[m]$ to $[n]$ is adjacency-preserving if it is strictly increasing for the product order and if $d((\epsilon\_1,\dots,\epsilon\_m),(\eta\_1,\dots,\eta\_m)) = 1$ implies $d(f(\epsilon\_1,\dots,\epsilon\_m),f(\eta\_1,\dots,\eta\_m)) = 1$. Example : the adjacency-preserving maps from $[2]$ to itself are $(x\_1,x\_2)\mapsto (x\_1,x\_2)$, $(x\_1,x\_2)\mapsto (x\_2,x\_1)$, $(x\_1,x\_2)\mapsto (\min(x\_1,x\_2),\max(x\_1,x\_2))$ and $(x\_1,x\_2) \mapsto (\max(x\_1,x\_2),\min(x\_1,x\_2))$.


Let $n\geq 3$. Let $f:[n]\to [n]$ be an adjacency-preserving map which commutes with all automorphisms of $[n]$ (the set of automorphisms of $[n]$ is in bijection with the permutation of the set $\{1,\dots,n\}$ by permuting the coordinates). Is $f$ necessarily the identity of $[n]$ ? For $n=2$, $(x\_1,x\_2)\mapsto (x\_2,x\_1)$ is a counter-example.


PS : the definitions are in this paper <http://www.pps.univ-paris-diderot.fr/~gaucher/symcub.pdf> (published paper here <http://dx.doi.org/10.1016/j.tcs.2009.11.013> if you have access);



About the $n$-cube
Identify elements of $[n]$ with subests of an $n$-element set. $f$ has to be level preserving because a maximal chain must be sent to a chain. If $f$ commutes with permutations, then the image is symmetric, so the image of $f$ contains all singletons. $f$ must be the identity on sets of size $1$ since $S\_n$ has no center for $n\gt 2$. You can determine each set $S$ by the singletons in a chain from the empty set to $S$, and $f(S)$ contains the same singletons, so $f$ fixes each set for $n\gt 3$. 


