As a very special case, I claim that there is no regular topology on a countably infinite set such that the only contiuous surjection $f:X\rightarrow X$ such that is the identity map. If $X$ is countably and regular, then $X$ is a regular Lindelof space. Since every regular Lindelof space is paracompact, the space $X$ is paracompact and hence completely regular and even realcompact. If $X$ is compact, then $X$ is isomorphic to some countable ordinal $\alpha+1$. Therefore, the mapping $g:\alpha+1\rightarrow\alpha+1$ where $g(n)=n-1$ where $n$ is a finite non-zero ordinal and $g(\beta)=\beta$ where $\beta$ is infinite or zero is a continuous surjection. 

Now assume that $X$ is not compact. Recall that a space is compact if and only if it is realcompact and pseudocompact. Therefore, since $X$ is not compact but $X$ is realcompact, the space $X$ is not pseudocompact. Furthermore, the space $X$ is zero-dimensional by the following argument: if $U$ is a neighborhood of $x$ and $f:X\rightarrow[0,1]$ is a mapping such that $f(x)=1$ and $f=0$ outside $U$, then the function $f$ is not surjective since $X$ is countable. Therefore, if $r\not\in f[X]$, then $f^{-1}[r,1]$ is a clopen set with
$x\in f^{-1}[r,1]\subseteq U$. Therefore, since $X$ is zero-dimensional but not pseudocompact, by [this answer][1], there is a partition of $X$ into infinitely many clopen sets $(C_{n})_{n\in\mathbb{N}}$. Let $(x_{n})_{n\in\mathbb{N}}$ be an enumeration of the elements of $X$. Then let $f:X\rightarrow X$ be the mapping where we let $f(x)=x_{n}$ whenever $x\in C_{n}$. Then $f$ is generally a non-identity continuous surjection.

I conjecture that it is consistent with the negation of the continuum hypothesis that every completely regular space of cardinality below the continuum has a non-identity surjection.


  [1]: http://mathoverflow.net/a/103644/22277