Spaces for which all transpositions are homeomorphisms The discrete, indiscrete, cofinite, cocountable, co-(any downwards closed collection of cardinals) topological spaces are part of the unique class of spaces such that all permutations of the space (bijections from the space to itself) are autohomeomorphisms. This is because their topologies may be defined purely in terms of cardinality, so everything is bijection-invariant. If we relax the requirement to just saying that all transpositions (i.e. permutations that just swap two elements) are homeomorphisms, do we get a broader class of spaces, or is this property equivalent to bijection-invariance?

Let's call a topological space $X$ [weakly] anti-rigid, if every permutation [transposition] is a homeomorphism.
And filter like, if whenever $U$ is a non-empty open subset of $X$, and $U \subset V \subset X$, then also $V$ is open.
Then we have:
$X$ weakly anti-rigid, iff $X$ is filter like and ($X$ is T1 or $X$ is indiscrete).
Proof. $"\Rightarrow"$: If $X$ contains an isolated point, then $X$ is discrete, hence filter like and T1. So let's assume that $X$ contains no isolated points, and $|X| \ge 2$.
Let $\emptyset \neq U$ be open, $y \in X$. Case 1: $y \in U$: then $U \cup \{y\} = U$ is open. Case 2: $y \notin U$. Choose an $x \in U$. Then the transposition
$f: X \rightarrow X$ swapping $x$ and $y$ shows that $(U \setminus \{x\}) \cup \{y\} = f(U)$ is open. Hence
$U \cup \{y\} = U \cup ((U \setminus \{x\}) \cup \{y\})$ is open.
Thus, any subset $V$ such that $U \subset V$ is open.
Now assume $X$ is not T1: then the intersection of all non-empty open sets is non-empty, hence equals $X$, hence X is indiscrete.

$"\Leftarrow"$: W.l.o.g. $X$ T1. Let $x, y \in X$,
$f: X \rightarrow X$ be the transposition swapping $x$ and $y$.
We have to show that $f$ is continuous: Let $U$ be open in $X$.
Case 1: $x, y \in U$ or $x, y \notin U$: Then $f^{-1} (U) = U$.
Case 2: $x \in U, y \notin U$: Then, by T1, $U \setminus \{x\}$ is open, hence $f^{-1}(U) =  (U \setminus \{x\}) \cup \{y\}$ is open.
Case 3: $x \notin U, y \in U$: analogous.
Now let $\mathfrak{F}$ be a free ultrafilter on $\mathbb N$.
Then $\mathfrak{F} \cup \{\emptyset \}$ is filter like topology on $\mathbb N$ and it is $T1$, hence weakly anti-rigid, but not anti-rigid:
Let $\mathbb N = A \cup B, A \cap B = \emptyset$, $A, B$ be infinite. W.l.o.g. $A \in \mathfrak{F}$.
There is a bijective map $f: \mathbb N \rightarrow \mathbb N$, such that $f(B) = A$. Hence, $f$ is not continuous.
Remark: it is easy to see that weakly anti-rigid Hausdorff spaces are discrete.
