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Let $S(X)$ be the group of permutations of $X$, $X$ infinite. It is a classical consequence of the Baer theorem (Onofri for $X$ countable) that $S(X)$ has no nontrivial finite quotient [actually every nontrivial quotient has cardinal $2^{|X|}$]. Hence every finite orbit of $S(X)$ on $T(X)$, the set of topologies on $X$, is a singleton. [Edit: actually $S(X)$ has no proper subgroup of index $<|X|$ and hence this even applies to orbits of cardinal $<|X|$.]

The orbits of $S(X)$ on $2^X$ are indexed by pairs $(u,v)$ of cardinals such that $\max(u,v)=|X|$, namely $C(u,v)$ the set of subsets of cardinal $u$ with complement of cardinal $v$. Let $W$ be the set of such pairs $(u,v)$; it has an obvious total ordering.

Hence an $S(X)$-invariant subset $\tau$ of $2^X$ is determined by a subset $E_\tau$ of $W$. The condition that $\tau$ is stable under taking arbitrary unions means that $(u,v)\le (u',v')$ and $(0,|X|)\neq (u,v)\in E_\tau$ implies $(u',v')\in E_\tau$. In addition, the condition of being stable under taking finite intersection means that $(u,v)\in E_\tau$, $v=|X|$ implies $(u',v')\in E_\tau$ whenever $(u,v)\le (u,v)$, and $(|X|,n)\in E_\tau$ for $0<n<\omega$ implies $(|X|,n')\in E_\tau$ for all $n'\ge n$.

We deduce that every $S(X)$-invariant topology on $X$ is one of the following

1. the indiscrete topology $\{\emptyset,X\}$;
2. for some infinite $\alpha\le |X|$, the topology $\tau_\alpha$ for which open subsets are $\emptyset$ and subsets with complement of cardinal $<\alpha$;
3. the discrete topology (this is the only Hausdorff one).

Note: this is equivalent to ask what are the topologies on $X$ for which the self-homeomorphism group is the whole group of permutations.

Let $S(X)$ be the group of permutations of $X$, $X$ infinite. It is a classical consequence of the Baer theorem (Onofri for $X$ countable) that $S(X)$ has no nontrivial finite quotient [actually every nontrivial quotient has cardinal $2^{|X|}$]. Hence every finite orbit of $S(X)$ on $T(X)$, the set of topologies on $X$, is a singleton. [Edit: actually $S(X)$ has no proper subgroup of index $<|X|$ and hence this even applies to orbits of cardinal $<|X|$.]

The orbits of $S(X)$ on $2^X$ are indexed by pairs $(u,v)$ of cardinals such that $\max(u,v)=|X|$, namely $C(u,v)$ the set of subsets of cardinal $u$ with complement of cardinal $v$. Let $W$ be the set of such pairs $(u,v)$; it has an obvious total ordering.

Hence an $S(X)$-invariant subset $\tau$ of $2^X$ is determined by a subset $E_\tau$ of $W$. The condition that $\tau$ is stable under taking arbitrary unions means that $(u,v)\le (u',v')$ and $(0,|X|)\neq (u,v)\in E_\tau$ implies $(u',v')\in E_\tau$. In addition, the condition of being stable under taking finite intersection means that $(u,v)\in E_\tau$, $v=|X|$ implies $(u',v')\in E_\tau$ whenever $(u',v')\le (u,v)$, and $(|X|,n)\in E_\tau$ for $0<n<\omega$ implies $(|X|,n')\in E_\tau$ for all $n'\ge n$.

We deduce that every $S(X)$-invariant topology on $X$ is one of the following

1. the indiscrete topology $\{\emptyset,X\}$;
2. for some infinite $\alpha\le |X|$, the topology $\tau_\alpha$ for which open subsets are $\emptyset$ and subsets with complement of cardinal $<\alpha$;
3. the discrete topology (this is the only Hausdorff one).

Note: this is equivalent to ask what are the topologies on $X$ for which the self-homeomorphism group is the whole group of permutations.

Let $S(X)$ be the group of permutations of $X$, $X$ infinite. It is a classical consequence of the Behr theorem (Onofri for $X$ countable) that $S(X)$ has no nontrivial quotient. Hence every finite orbit of $S(X)$ on $T(X)$, the set of topologies on $X$, is a singleton.

The orbits of $S(X)$ on $2^X$ are indexed by pairs $(u,v)$ of cardinals such that $\max(u,v)=|X|$, namely $C(u,v)$ the set of subsets of cardinal $u$ with complement of cardinal $v$. Let $W$ be the set of such pairs $(u,v)$; it has an obvious total ordering.

Hence an $S(X)$-invariant subset $\tau$ of $2^X$ is determined by a subset $E_\tau$ of $W$. The condition that $\tau$ is stable under taking arbitrary unions means that $(u,v)\le (u',v')$ and $(0,|X|)\neq (u,v)\in E_\tau$ implies $(u',v')\in E_\tau$. In addition, the condition of being stable under taking finite intersection means that $(u,v)\in E_\tau$, $v=|X|$ implies $(u',v')\in E_\tau$ whenever $(u,v)\le (u,v)$, and $(|X|,n)\in E_\tau$ for $0<n<\omega$ implies $(|X|,n')\in E_\tau$ for all $n'\ge n$.

We deduce that every $S(X)$-invariant topology on $X$ is one of the following

1. the indiscrete topology $\{\emptyset,X\}$;
2. for some infinite $\alpha\le |X|$, the topology $\tau_\alpha$ for which open subsets are $\emptyset$ and subsets with complement of cardinal $<\alpha$;
3. the discrete topology (this is the only Hausdorff one).

Note: this is equivalent to ask what are the topologies on $X$ for which the self-homeomorphism group is the whole group of permutations.

Let $S(X)$ be the group of permutations of $X$, $X$ infinite. It is a classical consequence of the Baer theorem (Onofri for $X$ countable) that $S(X)$ has no nontrivial finite quotient [actually every nontrivial quotient has cardinal $2^{|X|}$]. Hence every finite orbit of $S(X)$ on $T(X)$, the set of topologies on $X$, is a singleton. [Edit: actually $S(X)$ has no proper subgroup of index $<|X|$ and hence this even applies to orbits of cardinal $<|X|$.]

The orbits of $S(X)$ on $2^X$ are indexed by pairs $(u,v)$ of cardinals such that $\max(u,v)=|X|$, namely $C(u,v)$ the set of subsets of cardinal $u$ with complement of cardinal $v$. Let $W$ be the set of such pairs $(u,v)$; it has an obvious total ordering.

Hence an $S(X)$-invariant subset $\tau$ of $2^X$ is determined by a subset $E_\tau$ of $W$. The condition that $\tau$ is stable under taking arbitrary unions means that $(u,v)\le (u',v')$ and $(0,|X|)\neq (u,v)\in E_\tau$ implies $(u',v')\in E_\tau$. In addition, the condition of being stable under taking finite intersection means that $(u,v)\in E_\tau$, $v=|X|$ implies $(u',v')\in E_\tau$ whenever $(u,v)\le (u,v)$, and $(|X|,n)\in E_\tau$ for $0<n<\omega$ implies $(|X|,n')\in E_\tau$ for all $n'\ge n$.

We deduce that every $S(X)$-invariant topology on $X$ is one of the following

1. the indiscrete topology $\{\emptyset,X\}$;
2. for some infinite $\alpha\le |X|$, the topology $\tau_\alpha$ for which open subsets are $\emptyset$ and subsets with complement of cardinal $<\alpha$;
3. the discrete topology (this is the only Hausdorff one).

Note: this is equivalent to ask what are the topologies on $X$ for which the self-homeomorphism group is the whole group of permutations.

Let $S(X)$ be the group of permutations of $X$, $X$ infinite. It is a classical consequence of the Behr theorem (Onofri for $X$ countable) that $S(X)$ has no nontrivial quotient. Hence every finite orbit of $S(X)$ on $T(X)$, the set of topologies on $X$, is a singleton.

The orbits of $S(X)$ on $2^X$ are indexed by pairs $(u,v)$ of cardinals such that $\max(u,v)=|X|$, namely $C(u,v)$ the set of subsets of cardinal $u$ with complement of cardinal $v$. Let $W$ be the set of such pairs $(u,v)$; it has an obvious total ordering.

Hence an $S(X)$-invariant subset $\tau$ of $2^X$ is determined by a subset $E_\tau$ of $W$. The condition that $\tau$ is stable under taking arbitrary unions means that $(u,v)\le (u',v')$ and $(0,|X|)\neq (u,v)\in E_\tau$ implies $(u',v')\in E_\tau$. In addition, the condition of being stable under taking finite intersection means that $(u,v)\in E_\tau$, $v=|X|$ implies $(u',v')\in E_\tau$ whenever $(u,v)\le (u,v)$, and $(|X|,n)\in E_\tau$ for $0<n<\omega$ implies $(|X|,n')\in E_\tau$ for all $n'\ge n$.

We deduce that every $S(X)$-invariant topology on $X$ is one of the following

1. the indiscrete topology $\{\emptyset,X\}$;
2. for some infinite $\alpha\le |X|$, the topology $\tau_\alpha$ for which open subsets are $\emptyset$ and subsets with complement of cardinal $<\alpha$;
3. the discrete topology (this is the only Hausdorff one).

Let $S(X)$ be the group of permutations of $X$, $X$ infinite. It is a classical consequence of the Behr theorem (Onofri for $X$ countable) that $S(X)$ has no nontrivial quotient. Hence every finite orbit of $S(X)$ on $T(X)$, the set of topologies on $X$, is a singleton.

The orbits of $S(X)$ on $2^X$ are indexed by pairs $(u,v)$ of cardinals such that $\max(u,v)=|X|$, namely $C(u,v)$ the set of subsets of cardinal $u$ with complement of cardinal $v$. Let $W$ be the set of such pairs $(u,v)$; it has an obvious total ordering.

Hence an $S(X)$-invariant subset $\tau$ of $2^X$ is determined by a subset $E_\tau$ of $W$. The condition that $\tau$ is stable under taking arbitrary unions means that $(u,v)\le (u',v')$ and $(0,|X|)\neq (u,v)\in E_\tau$ implies $(u',v')\in E_\tau$. In addition, the condition of being stable under taking finite intersection means that $(u,v)\in E_\tau$, $v=|X|$ implies $(u',v')\in E_\tau$ whenever $(u,v)\le (u,v)$, and $(|X|,n)\in E_\tau$ for $0<n<\omega$ implies $(|X|,n')\in E_\tau$ for all $n'\ge n$.

We deduce that every $S(X)$-invariant topology on $X$ is one of the following

1. the indiscrete topology $\{\emptyset,X\}$;
2. for some infinite $\alpha\le |X|$, the topology $\tau_\alpha$ for which open subsets are $\emptyset$ and subsets with complement of cardinal $<\alpha$;
3. the discrete topology (this is the only Hausdorff one).

Note: this is equivalent to ask what are the topologies on $X$ for which the self-homeomorphism group is the whole group of permutations.