Hausdorff spaces with trivial automorphism group Is the singleton space the only Hausdorff space $X$ such that the set of automorphisms $\varphi: X\to X$ equals $\{\textrm{id}_X\}$?
 A: It's easy (if a little tedious, just a little) to construct a non-trivial metric dendroid (like an infinite tree or precisely--a one-dimensional 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 interval--that'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 end-point 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 set-theoretically 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.

(Non-triviality of a dendroid means, by definition, that it has more than one point).

The above dendroid (like any non-trivial dendroid) has non-constant 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.
A: 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 (non-trivial) 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, 201-208). I also recommend "Decompositions of rigid spaces" by van Engelen and van Mill (PAMS 89, 1983, 533-536), 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.
