Is the singleton space the only Hausdorff space $X$ such that the set of automorphisms $\varphi: X\to X$ equals $\{\textrm{id}_X\}$?

4$\begingroup$ The empty space is another example. $\;$ $\endgroup$– user5810Commented Dec 3, 2014 at 10:57

4$\begingroup$ There is also a series of papers by Kannan and Rajagopalan on the topic of rigid spaces. You can find there some further constructions and references. They also study strongly rigid spaces, which you mention in another question. $\endgroup$– Martin SleziakCommented Dec 3, 2014 at 15:47

$\begingroup$ The answer of Van Name contains some spaces whose morphism is very small mathoverflow.net/questions/198704/… $\endgroup$– Ali TaghaviCommented Feb 8, 2016 at 12:21
2 Answers
Not at all. Those spaces are called rigid and there are plenty of examples in the literature. The opposite notion is homogeneity which is a better studied property. The first (nontrivial) rigid space was constructed by Kuratowski in "Sur la puissance de l'ensemble des nombres de dimension au sens de M Frechet" (Fund.Math 8, 1926, 201208). I also recommend "Decompositions of rigid spaces" by van Engelen and van Mill (PAMS 89, 1983, 533536), where they give two nice examples: a rigid space that can be decomposed into two homeomorphic homogeneous subspaces and a homogeneous space that can be decomposed into two homeomorphic rigid subspaces. By the way, these two examples are even separable and metrizable.
It's easy (if a little tedious, just a little) to construct a nontrivial metric dendroid (like an infinite tree or preciselya onedimensional compact metric AR) such that
 the ramification degree of each point is finite;
 for each natural number $\ n=1\ 2\ \ldots\ $ there exists exactly one point which has ramification $\ n+2$;
 the set of all points of ramification $\ > 2\ $ is dense.
Then for every homeomorphism of our dendroid onto itself each point which has ramification $\ > 2\ $ is fixed. It follows that every such homeomorphism is the identity of our dendroid.
A construction is obtained by starting with an intervalthat's your initial stage of the induction, then at each stage add several short intervals which have one end at the middle of each existing intervals; at the center of $n$th interval you add $\ n\ $ interval so that ramification will be the unique $\ n+2.\ $ You iterate this at infinitum. You folow it up with ending the endpoint at each combinatorially infinite branch (but their geodesic metric length is finite), thus we get a compact.
REMARK An interval of a tree is meant any maximal interval such that all its internal points have ramification $\ 2.\ $ Each interval of a tree at any next stage, which was settheoretically contained in the previous tree, is one of the two halves of the respective larger interval of that previous tree. Regardless of the stage at which theses intervals occur, they are all numbered $\ 1\ 2\ \ldots$. With each stage there is a bunch of intervals which are consecutively indexed above the previous tree, and below the next one.
(Nontriviality of a dendroid means, by definition, that it has more than one point).
The above dendroid (like any nontrivial dendroid) has nonconstant continuous maps into itself, different from identity. Indeed, this is true for arbitrary AR which has more than one point. Thus the given example, while rigid, is not strongly rigid.

2$\begingroup$ Sorry, I’m confused. How is “has more than one autohomeomorphism” different from “not rigid” as defined in the question? $\endgroup$ Commented Dec 3, 2014 at 21:00

$\begingroup$ Sorry, it was just a (big) TYPO as opposed simply to a typo. I have replaced it with correct text. Thank you @Emil for pointing it out to me. $\endgroup$ Commented Dec 3, 2014 at 22:19

$\begingroup$ I see, thanks for the clarification. $\endgroup$ Commented Dec 3, 2014 at 22:25