DEF. Let $X, C$ be topological spaces. $X$ is said to be $n$-fold transitive with respect to $C$ if for every collection of $n$ injective continuous maps $\iota_i\colon C \to X$ with disjoint image and every permutation $\sigma\in S_n$ we find an automorphism $h\colon X\to X$ satisfying $h\circ \iota_i = \iota_{\sigma(i)}$.
Here, the $C$ is meant to stand for “characteristic”. Note that this is quite a strong statement: Even if we take $C$ to be the simplest topological space, namely the point, even only 2-fold transitivity appears pretty strong: every point can be transformed into every other, implying there must be a lot of automorphisms (read: symmetry) intrinsic to $X$. However, it might be of interest to study maximal n-fold subspaces. Also, keep in mind that n-fold transitivity implies $n-1$-fold transitivity by considering a permutation that leaves one image fixed – which matches the suspicion that transitivity for higher $n$ represents a stronger property.
Q. Can a space be 2-fold with respect to $S$, but not a subspace $T\subseteq S$? A. Yes. Consider the “converging sequence” space $$\newcommand{\C}{\mathscr C} \left\{\frac{1}{n}\mid n\in \mathbb N\right\}\cup \{0\} =: \C, $$ and take two copies of that: $X := \C \sqcup \C$. This is 2-transitive w.r.t. $\C$ (check that!), but not w.r.t. the one-point space $\ast$, since the two points $(0, 0)$ and $(0, \frac{1}{2})$ (the first component denotes the coloring introduced by our disjoint union) cannot be interchanged by a homeomorphism, since one point is isolated, and the other is not.
This appears weird, but is actually quite interesting in the following way: being n-fold transitive with respect to the one-point space is a pretty harsh restriction already, but perhaps by choosing „wisely“, we can still draw information about the symmetry of other spaces we wouldn't have been able to “see” by just considering the continuous permutability of points.
From here on, if we make no mention of the embedded space, we mean to analyze the one-point case.
Ex. $\mathbb R$ is 2-transitive, but not 3-transitive: There is no homeomorphism taking $(-1,0,1)$ to $(-1,1,0)$, as every homeo must preserve the usual order (wait, is that true?)
Ex. $\DeclareMathOperator{cofin}{CoFin}X:=\cofin(S)$ for an infinite set $S$ is n-transitive for any $n$ and any $C$, because every bijection $S\to S$ is a homeomorphism: For bijections, preimages of finite sets are finite, therefore, preimages of cofinite sets must be cofinite. Given a permutation of images of $C$ in $S$, the map that permutes the images and acts as the identity on everything else constitutes a valid choice of such an automorphism.
Q. Is the cofinite space the only one with this property (fixing $C$, demanding arbitrary high $n$)?
Note. When constructing examples by taking coproducts, keep in mind that once we can partition our space into (finitely many) connected components, every automorphism must be a permutation of those connected components, implying by restriction that they may only mediate between connected components which are homeomorphic.
Conjecture. If $X$ is not n-transitive w.r.t. $C$, it is probably not $n$-transitive w.r.t. $C\sqcup C$.
Example. Given any $n\geq 2$, a subpace of $\C$ is $n$-transitive ($n\geq 2$ implies it must contain at least two elements) if and only if it does not contain $0$. Therefore, there is no maximal $n$-transitive subspace. So let's try to make the characteristic space more complicated: finite or larger discrete spaces probably won't behave any differently, as they're essentially just a bunch of points, but let's look at $\C$ itself, which is the “next more complicated“ example (it has only one non-isolated point). Unfortunately though, we can only inject one copy of $\C$ into $\C$ at a time, given that the non-isolatedness of $0$ must be preserved. So $n\geq 2$-transitivity is impossible as well.
Q. Are there non-discrete characteristical spaces which give rise to non-trivial transitivity of $\C$? A. Let's look at $\operatorname{Aut}(\C)$ explicitly: Every such automorphism $h$ must preserve $0$, ergo be a permutation n the rest. But which permutations are allowed? Well, every open set containing 0 is of the form $[0,1/k]\cup I$, where $I$ is an arbitrary subset of points in $C$ outside of said interval (the interval is to be interpreted not as a subset of $\mathbb R$, but of the poset $\C$). Put another way, they are precisely the cofinite sets containing 0. The property of being cofinite is clearly preserved if $h$ is a bijection, and as it also preserves 0, the preimage is open without need for further demands. So indeed, any permutation leaving 0 constant constitutes a valid automorphism of $\C$. However, a pretty basic problem cannot be resolved: Anything that's not countable does not admit an injection into $\C$, and nor does anything with more than one non-isolated point. The only countable space with one isolated point is $\C$(?), so lets pick the space with no isolated points: the discrete space $\mathscr P (\mathbb N)$. However, injecting this space does not even allow for 2-transitivity, because again, one injection might containg $0$ in its image and another might not, and both are not translatable by an automorphism. So $\C$ does not admit any transitivity, and may rightfully be proclaimed totally asymmetric.
Pre-Proof (the only ctbl space with one isolated point is $\C$): idea: embed $\C$ injectively into this space (possible, as the poset of open sets containing the non-isolated points does not have a minimal element, apply choice to this chain of decreasing open sets), and note that the complement consists of isolated points and is thus discrete. The complement can be probably made finite by choosing a maximal chain (note that every open set can “constructively” be extended by gluing a point), so we have hopefully our image is an open set