Joel's answer is a special case of the following. Let G be a transitive permutation group on a finite topological space X with more than 2 elements. Then G together with constants is never the whole monoid of all cts self maps of X.
Proof. A finite topological space is just a finite preordered set. If the preorder has all elements equivalent we have the indiscrete topology and so all maps are cts. If the preorder is equality then the topology is discrete and so all maps are cts.
Suppose $x\leq y$ and $gx=y$. 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. This the preorder is an equivalence relation. If there is more than 1 class and 1 class is not a singleton, then crushing the non-singleton is cts but not in the monoid. So the preorder is equality or the universal equivalence relation.