Joel's answer is a special case of the following. Let G be a transitive permutation group of continuous maps on a finite topological space X with more than 2 elements. Then G together with all constants is never the whole monoid of all continuous self maps of X.
Proof. A finite topological space is just a finite preordered set via the specialization ordering/Alexandrov topology. Continuous maps are precisely order preserving maps.
If the preorder is the universal equivalence relation we have the indiscrete topology and so all maps are continuous. If the preorder is equality then the topology is discrete and so all maps are continuous. Since there are always maps on a three or more element set which are neither constant nor permutations we are done in these two cases. Let's prove only these two cases occur.
Suppose that $x\leq y$ and $gx=y$ with $g\in G$ by transitivity. Then $x\leq gx$ and so $x\leq gx\leq g^2x\leq \cdots$. Since G is finite we eventually get $g^n=1$ and so $y=gx\leq x$. Thus any comparable elements are equivalent. Hence the preorder is an equivalence relation. If there is more than 1 class and also some class is not a singleton, then crushing the non-singleton is continuous but not in the monoid. So the preorder is equality or the universal equivalence relation.