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Yes, such a mapping necessarily sends singletons to singletons.

Let $f\colon\mathcal P(S)\to\mathcal P(T)$ be a monotone bijection. By monotony, $f(\varnothing)\subseteq f(X)\subseteq f(S)$ for all $X\subseteq S$, hence using the surjectivity of $f$, $f(\varnothing)=\varnothing$ and $f(S)=T$. Also, as noted in the question, $f$-inverses of singletons must be singletons: if $f(X)=\{t\}$, we must have $X\ne\varnothing$; if we further assume for contradiction that there exists $\varnothing\ne X'\subsetneq X$, then $\varnothing\ne f(X')\subsetneq\{t\}$, which is impossible.

Thus, there is $S'\subseteq S$ and a bijection $g\colon S'\to T$ such that $f(\{s\})=\{g(s)\}$ for all $s\in S'$. But then by monotony, $\{t\}\subseteq f(S')$ for all $t\in\operatorname{im}(g)=T$, i.e., $f(S')=T=f(S)$, which implies $S'=S$.

Furthermore, we have $f(X)=g[X]$ for all $X\subseteq S$, hence $f$ is an order isomorphism. To see this, note that for any $s\in S$, $f(S\smallsetminus\{s\})$ includes $f(\{s'\})=\{g(s')\}$ for all $s'\ne s$, but it is strictly contained in $T=f(S)$, hence $f(S\smallsetminus\{s\})=T\smallsetminus\{g(s)\}$. Then for any $X\subseteq S$, we have $\{g(s)\}\subseteq f(X)\subseteq T\smallsetminus\{g(s')\}$ for all $s\in X$ and $s'\notin X$, whence $f(X)=g[X]$.

Yes, such a mapping necessarily sends singletons to singletons.

Let $f\colon\mathcal P(S)\to\mathcal P(T)$ be a monotone bijection (or more generally, a surjective strictly monotone function). By monotony, $f(\varnothing)\subseteq f(X)\subseteq f(S)$ for all $X\subseteq S$, hence using the surjectivity of $f$, $f(\varnothing)=\varnothing$ and $f(S)=T$. Also, as noted in the question, $f$-inverses of singletons must be singletons: if $f(X)=\{t\}$, we must have $X\ne\varnothing$; if we further assume for contradiction that there exists $\varnothing\ne X'\subsetneq X$, then $\varnothing\ne f(X')\subsetneq\{t\}$, which is impossible.

Thus, there is $S'\subseteq S$ and a bijection $g\colon S'\to T$ such that $f(\{s\})=\{g(s)\}$ for all $s\in S'$. But then by monotony, $\{t\}\subseteq f(S')$ for all $t\in\operatorname{im}(g)=T$, i.e., $f(S')=T=f(S)$, which implies $S'=S$.

Furthermore, we have $f(X)=g[X]$ for all $X\subseteq S$, hence $f$ is an order isomorphism. To see this, note that for any $s\in S$, $f(S\smallsetminus\{s\})$ includes $f(\{s'\})=\{g(s')\}$ for all $s'\ne s$, but it is strictly contained in $T=f(S)$, hence $f(S\smallsetminus\{s\})=T\smallsetminus\{g(s)\}$. Then for any $X\subseteq S$, we have $\{g(s)\}\subseteq f(X)\subseteq T\smallsetminus\{g(s')\}$ for all $s\in X$ and $s'\notin X$, whence $f(X)=g[X]$.

Yes, such a mapping necessarily sends singletons to singletons.

Let $f\colon\mathcal P(S)\to\mathcal P(T)$ be a monotone bijection. By monotony, $f(\varnothing)\subseteq f(X)\subseteq f(S)$ for all $X\subseteq S$, hence using the surjectivity of $f$, $f(\varnothing)=\varnothing$ and $f(S)=T$. Also, as noted in the question, $f$-inverses of singletons must be singletons: if $f(X)=\{t\}$, we must have $X\ne\varnothing$; if we further assume for contradiction that there exists $\varnothing\ne X'\subsetneq X$, then $\varnothing\ne f(X')\subsetneq\{t\}$, which is impossible.

Thus, there is $S'\subseteq S$ and a bijection $g\colon S'\to T$ such that $f(\{s\})=\{g(s)\}$ for all $s\in S'$. But then by monotony, $\{t\}\subseteq f(S')$ for all $t\in\operatorname{im}(g)=T$, i.e., $f(S')=T=f(S)$, which implies $S'=S$.

Furthermore, we have $f(X)=g[X]$ for all $X\subseteq S$, hence $f$ is an order isomorphism. To see this, note that for any $s\in S$, $f(S\smallsetminus\{s\})$ includes $f(\{s'\})=\{g(s')\}$ for all $s'\ne s$, but it is strictly contained in $T=f(S)$, hence $f(S\smallsetminus\{s\})=T\smallsetminus\{g(s)\}$. Then for any $X\subseteq S$, we have $\{g(s)\}\subseteq f(X)\subseteq T\smallsetminus\{g(s')\}$ for all $s\in X$ and $s'\notin S$, whence $f(X)=g[X]$.

Yes, such a mapping necessarily sends singletons to singletons.

Let $f\colon\mathcal P(S)\to\mathcal P(T)$ be a monotone bijection. By monotony, $f(\varnothing)\subseteq f(X)\subseteq f(S)$ for all $X\subseteq S$, hence using the surjectivity of $f$, $f(\varnothing)=\varnothing$ and $f(S)=T$. Also, as noted in the question, $f$-inverses of singletons must be singletons: if $f(X)=\{t\}$, we must have $X\ne\varnothing$; if we further assume for contradiction that there exists $\varnothing\ne X'\subsetneq X$, then $\varnothing\ne f(X')\subsetneq\{t\}$, which is impossible.

Thus, there is $S'\subseteq S$ and a bijection $g\colon S'\to T$ such that $f(\{s\})=\{g(s)\}$ for all $s\in S'$. But then by monotony, $\{t\}\subseteq f(S')$ for all $t\in\operatorname{im}(g)=T$, i.e., $f(S')=T=f(S)$, which implies $S'=S$.

Furthermore, we have $f(X)=g[X]$ for all $X\subseteq S$, hence $f$ is an order isomorphism. To see this, note that for any $s\in S$, $f(S\smallsetminus\{s\})$ includes $f(\{s'\})=\{g(s')\}$ for all $s'\ne s$, but it is strictly contained in $T=f(S)$, hence $f(S\smallsetminus\{s\})=T\smallsetminus\{g(s)\}$. Then for any $X\subseteq S$, we have $\{g(s)\}\subseteq f(X)\subseteq T\smallsetminus\{g(s')\}$ for all $s\in X$ and $s'\notin X$, whence $f(X)=g[X]$.

Yes, such a mapping necessarily sends singletons to singletons.

Let $f\colon\mathcal P(S)\to\mathcal P(T)$ be a monotone bijection. As noted in the question, $f$-inverses of singletons must be singletons, hence there is $S'\subseteq S$ and a bijection $g\colon S'\to T$ such that $f(\{s\})=\{g(s)\}$ for all $s\in S'$. But then by monotony, $\{t\}\subseteq f(S')$ for all $t\in\operatorname{im}(g)=T$, i.e., $f(S')=T$, which implies $S'=S$.

Furthermore, we have $f(X)=g[X]$ for all $X\subseteq S$, hence $f$ is an order isomorphism. To see this, note that for any $s\in S$, $f(S\smallsetminus\{s\})$ includes $f(\{s'\})=\{g(s')\}$ for all $s'\ne s$, but it is strictly contained in $T=f(S)$, hence $f(S\smallsetminus\{s\})=T\smallsetminus\{g(s)\}$. Then for any $X\subseteq S$, we have $\{g(s)\}\subseteq f(X)\subseteq T\smallsetminus\{g(s')\}$ for all $s\in X$ and $s'\notin S$, whence $f(X)=g[X]$.

Yes, such a mapping necessarily sends singletons to singletons.

Let $f\colon\mathcal P(S)\to\mathcal P(T)$ be a monotone bijection. By monotony, $f(\varnothing)\subseteq f(X)\subseteq f(S)$ for all $X\subseteq S$, hence using the surjectivity of $f$, $f(\varnothing)=\varnothing$ and $f(S)=T$. Also, as noted in the question, $f$-inverses of singletons must be singletons: if $f(X)=\{t\}$, we must have $X\ne\varnothing$; if we further assume for contradiction that there exists $\varnothing\ne X'\subsetneq X$, then $\varnothing\ne f(X')\subsetneq\{t\}$, which is impossible.

Thus, there is $S'\subseteq S$ and a bijection $g\colon S'\to T$ such that $f(\{s\})=\{g(s)\}$ for all $s\in S'$. But then by monotony, $\{t\}\subseteq f(S')$ for all $t\in\operatorname{im}(g)=T$, i.e., $f(S')=T=f(S)$, which implies $S'=S$.

Furthermore, we have $f(X)=g[X]$ for all $X\subseteq S$, hence $f$ is an order isomorphism. To see this, note that for any $s\in S$, $f(S\smallsetminus\{s\})$ includes $f(\{s'\})=\{g(s')\}$ for all $s'\ne s$, but it is strictly contained in $T=f(S)$, hence $f(S\smallsetminus\{s\})=T\smallsetminus\{g(s)\}$. Then for any $X\subseteq S$, we have $\{g(s)\}\subseteq f(X)\subseteq T\smallsetminus\{g(s')\}$ for all $s\in X$ and $s'\notin S$, whence $f(X)=g[X]$.