Is there a nonrigid space X homeomorphic to X/Aut(X)? Is there a topological space $X$ satisfying Conditions (a) and (b) below?
(a) $X$ has at least one nontrivial automorphism,
(b) $X$ is homeomorphic to the quotient $X/\!\operatorname{Aut}(X)$ of $X$ by the action of its automorphism group.
 A: Here is a general theorem that can be used to construct examples of such spaces.
Theorem: Let $X$ be a nonempty topological space.  Then there exists a topological space $Y$ such that $|Y|>|X|$ and a homeomorphism $Y/\operatorname{Aut}(Y)\to X$ which induces a bijection between the clopen subsets of $Y$ and the clopen subsets of $X$.  (In particular, if $X$ is connected, then so is $Y$.)
Given this theorem, start with a nonempty rigid connected space $X_0$ (say, a singleton).  Recursively use the theorem to obtain a sequence of connected spaces $X_n$ such that $X_n\cong X_{n+1}/\operatorname{Aut}(X_{n+1})$ and $|X_{n+1}|>|X_n|$.  Now let $X$ be the disjoint union of all these spaces $X_n$, together with an additional copy of $X_0$.  No two $X_n$'s are homeomorphic (since they all have different cardinalities), so any automorphism of $X$ must map each $X_n$ to itself (except that it can swap the two $X_0$'s).  So the quotient $X/\operatorname{Aut}(X)$ will turn $X_n$ into $X_{n-1}$ for each $n>0$ and will identify the two copies of $X_0$ together into one.  Thus, $X/\operatorname{Aut}(X)$ is a disjoint union one copy of each $X_n$ together with an extra copy of $X_0$, so it is homeomorphic to $X$.
Proof of Theorem: Consider the following general construction.  Suppose $(S_x)_{x\in X}$ is a family of nonempty totally ordered sets indexed by points of our space $X$.  We define a topological space $Y$ as follows.  The underlying set is the disjoint union $\bigsqcup_{x\in X} S_x$.  Let $p:Y\to X$ be the map sending each $S_x$ to $x$.  We topologize $Y$ by saying a set $U\subseteq Y$ is open if it satisfies the following two conditions:

*

*$p(U)$ is open in $X$.

*For all $x\in X$ and $a,b\in S_x$, if $a\leq b$ and $b\in U$, then $a\in U$.

In particular, for $A\subseteq X$, this means $A$ is open iff $p^{-1}(A)$ is open, so $p$ is a quotient map.  Also, each $S_x$ is connected as a subset of $Y$, so $p$ induces a bijection between the clopen subsets of $Y$ and the clopen subsets of $X$.
Now assume additionally that the totally ordered sets $S_x$ are all homogeneous, have more than one point, and are pairwise nonisomorphic.  I claim that then the orbits of the action of $\operatorname{Aut}(Y)$ on $Y$ are exactly the sets $S_x$, so that $p$ realizes $X$ as the quotient $Y/\operatorname{Aut}(Y)$.
First, if $f_x:S_x\to S_x$ is an order automorphism of $S_x$ for each $x$, then these maps combine to give an automorphism of $Y$.  So, each orbit of $\operatorname{Aut}(Y)$ is a union of sets $S_x$, since each $S_x$ is homogeneous.  Now note that for $a,b\in Y$, every open set containing $a$ contains $b$ iff $a$ and $b$ are in the same $S_x$ and $a\geq b$. (Here we use the assumption that every $S_x$ has more than one point to construct an open set containing $a$ but not $b$ if they are in different $S_x$'s.)  It follows that any automorphism of $Y$ must map each $S_x$ to some other $S_y$ by an order isomorphism.  But since the $S_x$'s are pairwise nonisomorphic, this means any automorphism must map each $S_x$ to itself.  Thus each $S_x$ is an orbit, as desired.
All that remains to be shown is that such a family $(S_x)$ of homogeneous non-isomorphic totally ordered sets actually exists (and moreover that such a family can be chosen such that $|Y|>|X|$).  This follows from the following lemma, which enables us to choose the $S_x$'s to all have different cardinalities which are as large as we like.
Lemma: For any cardinal $\kappa$, there exists a homogeneous totally ordered set of cardinality greater than $\kappa$.
Proof: Take $\mathbb{Z}^\kappa$ with the lexicographic order.  It is homogeneous since for any $a\in\mathbb{Z}^\kappa$, the map $b\mapsto b+a$ is an order-automorphism.
