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I think the question loses some of its appeal once we note that the preorder $T\sqsubseteq T'$ defined by the OP coincides with the preorder defined by the inclusion $\mathtt{Th}(T)\subseteq\mathtt{Th}(T')$ when restricted to theories which are strong enough to prove say $\exists!x\forall y(y\not\in x)$ (a formula suggested by Francois' comment). So if we restrict our analysis to theories which are closed under deduction ($\{\phi:T\vdash\phi\}=\mathtt{Th}(T)=T$), this preorder is simply the usual inclusion and if we focus on consistent theories (as Noah suggests) we are simply asking if a consistent theory can be extended to a maximal consistent theory (yes it can and it will be complete hence not recursively axiomatizable). So in the light of this, I think Noah's answer could be made simpler (This should really be a comment of mine, but my rep doesn't allow me to use that option). To see that $T\sqsubseteq T'$ is equivalent to $\mathtt{Th}(T)\subseteq\mathtt{Th}(T')$ (for strong enough theories), the hardest part is to focus on $\Rightarrow$. So assuming that $T\sqsubseteq T'$ and $\phi$ is a sentence such that $T\vdash \phi$, hence $T\vdash\exists !x\phi\land\forall y(y\not\in x)$ from which we obtain $T'\vdash\exists !x\phi\land\forall y(y\not\in x)$ and finally $T'\vdash\phi$.

I think the question loses some of its appeal once we note that the preorder $T\sqsubseteq T'$ defined by the OP coincides with the preorder defined by the inclusion $\mathtt{Th}(T)\subseteq\mathtt{Th}(T')$ when restricted to theories which are strong enough to prove say $\exists!x\forall y(y\not\in x)$ (a formula suggested by Francois' comment). So if we restrict our analysis to theories which are closed under deduction ($\{\phi:T\vdash\phi\}=\mathtt{Th}(T)=T$), this preorder is simply the usual inclusion and if we focus on consistent theories (as Noah suggests) we are simply asking if a consistent theory can be extended to a maximal consistent theory (yes it can and it will be complete hence not recursively axiomatizable). So in the light of this, I think Noah's answer could be made simpler (This should really be a comment of mine, but my rep doesn't allow me to use that option). To see that $T\sqsubseteq T'$ is equivalent to $\mathtt{Th}(T)\subseteq\mathtt{Th}(T')$ (for strong enough theories), the hardest part is to focus on $\Rightarrow$. So assuming that $T\sqsubseteq T'$ and $\phi$ is a sentence such that $T\vdash \phi$, we have $T\vdash\exists !x\phi\land\forall y(y\not\in x)$ from which we obtain $T'\vdash\exists !x\phi\land\forall y(y\not\in x)$ and finally $T'\vdash\phi$.

This feels reminiscent of Lindenbaum's lemma ascerting the existence of a maximal consistent set containing a given consistent set. I am wondering if something similar can be done here. First of all, I don't think $\sqsubseteq$ is a partial order unless we look the quotien set ${\cal F}/\sim$ modulo the equivalence $T\sim T'$ defined by $O_{n}(T)=O_{n}(T')$ for all $n$. Given a consistent set $T_{0}$ with equivalence class $[T_{0}]$ we may ask whether ${\cal F}=\{[T]: T\ \mbox{consistent},\, [T_{0}]\sqsubseteq[T]$} has a maximal element, and hope to apply Zorn's lemma. So we need to check that every chain in ${\cal F}$ has an upper bound in ${\cal F}$. Now if ${\cal C}\subseteq{\cal F}$ is a chain in ${\cal F}$ (we can assume it is not empty), it is tempting to define $T^{*}=\cup\{T: [T]\in{\cal C}\}$ and hope that $[T^{*}]$ is an upper bound (the set $T^{*}$ is not uniquely defined but hopefully $[T^{*}]$ is). Now let us fix a variable name '$x$' and consider the mapping $\tau:{\cal L}-\mbox{Form}\to{\cal L}-\mbox{Form}$ defined by $\tau(\phi)=\forall y_{1}\ldots\forall y_{n}\exists !x\phi$ where it is understood that $\{y_{1},\ldots,y_{n}\}=\mathtt{Fr}(\phi)\setminus{\{x\}}$ (free variables of $\phi$ except $x$). (so $\tau$ is a form of universal closure mapping but not quite). Then it seems to me the equivalence $T\sim T'$ can be expressed as $T\vdash \tau(\phi)\Leftrightarrow T'\vdash\tau(\phi)$ for all $\phi$. Denothing $\mathtt{Th}(T)=\{\phi:T\vdash\phi\}$, the equivalence becomes an equality between inverse images $\tau^{-1}(\mathtt{Th}(T))=\tau^{-1}(\mathtt{Th}(T'))$. Similarly the partial order $[T]\sqsubseteq[T']$ can be expressed as $\tau^{-1}(\mathtt{Th}(T))\subseteq\tau^{-1}(\mathtt{Th}(T'))$. Now coming back to our upper-bound candidate $[T^{*}]$ the key is that since ${\cal C}$ is totally ordered we have $\mathtt{Th}(T^{*})=\cup\{\mathtt{Th}(T): [T]\in{\cal C}\}$ and $\tau^{-1}(\mathtt{Th}(T^{*}))=\cup\{\tau^{-1}(\mathtt{Th}(T)):[T]\in{\cal C}\}$. Furthermore $T^{*}$ is consistent. So I have the feeling this works. Please forgive me if I am completely off the mark.

N.B. I should probably have pointed out that if $T\sim T'$ and $T$ is consistent, then $T'$ is also consistent. Otherwise, from $T'\vdash\bot$ we get $T'\vdash\exists! x(\bot)$ and so $T\vdash \exists! x(\bot)$ and finally $T\vdash\bot$.

N.B. I was worried for a second that the above argument would somehow be devoid of substance, by the fact that a maximal element $[T^{*}]$ could simply be obtained by taking the class modulo $\sim$ of a maximal consistent set $T^{*}\supseteq T_{0}$. This does not seem to be the case: $T^{*}$ may be maximal consistent. It does not mean it maximizes the number of mathematical objects it creates, so to speak.

I think the question loses some of its appeal once we note that the preorder $T\sqsubseteq T'$ defined by the OP coincides with the preorder defined by the inclusion $\mathtt{Th}(T)\subseteq\mathtt{Th}(T')$ when restricted to theories which are strong enough to prove say $\exists!x\forall y(y\not\in x)$ (a formula suggested by Francois' comment). So if we restrict our analysis to theories which are closed under deduction ($\{\phi:T\vdash\phi\}=\mathtt{Th}(T)=T$), this preorder is simply the usual inclusion and if we focus on consistent theories (as Noah suggests) we are simply asking if a consistent theory can be extended to a maximal consistent theory (yes it can and it will be complete hence not recursively axiomatizable). So in the light of this, I think Noah's answer could be made simpler (This should really be a comment of mine, but my rep doesn't allow me to use that option). To see that $T\sqsubseteq T'$ is equivalent to $\mathtt{Th}(T)\subseteq\mathtt{Th}(T')$ (for strong enough theories), the hardest part is to focus on $\Rightarrow$. So assuming that $T\sqsubseteq T'$ and $\phi$ is a sentence such that $T\vdash \phi$, hence $T\vdash\exists !x\phi\land\forall y(y\not\in x)$ from which we obtain $T'\vdash\exists !x\phi\land\forall y(y\not\in x)$ and finally $T'\vdash\phi$.

This feels reminiscent of Lindenbaum's lemma ascerting the existence of a maximal consistent set containing a given consistent set. I am wondering if something similar can be done here. First of all, I don't think $\sqsubseteq$ is a partial order unless we look the quotien set ${\cal F}/\sim$ modulo the equivalence $T\sim T'$ defined by $O_{n}(T)=O_{n}(T')$ for all $n$. Given a consistent set $T_{0}$ with equivalence class $[T_{0}]$ we may ask whether ${\cal F}=\{[T]: T\ \mbox{consistent},\, [T_{0}]\sqsubseteq[T]$} has a maximal element, and hope to apply Zorn's lemma. So we need to check that every chain in ${\cal F}$ has an upper bound in ${\cal F}$. Now if ${\cal C}\subseteq{\cal F}$ is a chain in ${\cal F}$ (we can assume it is not empty), it is tempting to define $T^{*}=\cup\{T: [T]\in{\cal C}\}$ and hope that $[T^{*}]$ is an upper bound (the set $T^{*}$ is not uniquely defined but hopefully $[T^{*}]$ is). Now let us fix a variable name '$x$' and consider the mapping $\tau:{\cal L}-\mbox{Form}\to{\cal L}-\mbox{Form}$ defined by $\tau(\phi)=\forall y_{1}\ldots\forall y_{n}\exists !x\phi$ where it is understood that $\{y_{1},\ldots,y_{n}\}=\mathtt{Fr}(\phi)\setminus{\{x\}}$ (free variables of $\phi$ except $x$). (so $\tau$ is a form of universal closure mapping but not quite). Then it seems to me the equivalence $T\sim T'$ can be expressed as $T\vdash \tau(\phi)\Leftrightarrow T'\vdash\tau(\phi)$ for all $\phi$. Denothing $\mathtt{Th}(T)=\{\phi:T\vdash\phi\}$, the equivalence becomes an equality between inverse images $\tau^{-1}(\mathtt{Th}(T))=\tau^{-1}(\mathtt{Th}(T'))$. Similarly the partial order $[T]\sqsubseteq[T']$ can be expressed as $\tau^{-1}(\mathtt{Th}(T))\subseteq\tau^{-1}(\mathtt{Th}(T'))$. Now coming back to our upper-bound candidate $[T^{*}]$ the key is that since ${\cal C}$ is totally ordered we have $\mathtt{Th}(T^{*})=\cup\{\mathtt{Th}(T): [T]\in{\cal C}\}$ and $\tau^{-1}(\mathtt{Th}(T^{*}))=\cup\{\tau^{-1}(\mathtt{Th}(T)):[T]\in{\cal C}\}$. Furthermore $T^{*}$ is consistent. So I have the feeling this works. Please forgive me if I am completely off the mark.

N.B. I should probably have pointed out that if $T\sim T'$ and $T$ is consistent, then $T'$ is also consistent. Otherwise, from $T'\vdash\bot$ we get $T'\vdash\exists! x(\bot)$ and so $T\vdash \exists! x(\bot)$ and finally $T\vdash\bot$.

This feels reminiscent of Lindenbaum's lemma ascerting the existence of a maximal consistent set containing a given consistent set. I am wondering if something similar can be done here. First of all, I don't think $\sqsubseteq$ is a partial order unless we look the quotien set ${\cal F}/\sim$ modulo the equivalence $T\sim T'$ defined by $O_{n}(T)=O_{n}(T')$ for all $n$. Given a consistent set $T_{0}$ with equivalence class $[T_{0}]$ we may ask whether ${\cal F}=\{[T]: T\ \mbox{consistent},\, [T_{0}]\sqsubseteq[T]$} has a maximal element, and hope to apply Zorn's lemma. So we need to check that every chain in ${\cal F}$ has an upper bound in ${\cal F}$. Now if ${\cal C}\subseteq{\cal F}$ is a chain in ${\cal F}$ (we can assume it is not empty), it is tempting to define $T^{*}=\cup\{T: [T]\in{\cal C}\}$ and hope that $[T^{*}]$ is an upper bound (the set $T^{*}$ is not uniquely defined but hopefully $[T^{*}]$ is). Now let us fix a variable name '$x$' and consider the mapping $\tau:{\cal L}-\mbox{Form}\to{\cal L}-\mbox{Form}$ defined by $\tau(\phi)=\forall y_{1}\ldots\forall y_{n}\exists !x\phi$ where it is understood that $\{y_{1},\ldots,y_{n}\}=\mathtt{Fr}(\phi)\setminus{\{x\}}$ (free variables of $\phi$ except $x$). (so $\tau$ is a form of universal closure mapping but not quite). Then it seems to me the equivalence $T\sim T'$ can be expressed as $T\vdash \tau(\phi)\Leftrightarrow T'\vdash\tau(\phi)$ for all $\phi$. Denothing $\mathtt{Th}(T)=\{\phi:T\vdash\phi\}$, the equivalence becomes an equality between inverse images $\tau^{-1}(\mathtt{Th}(T))=\tau^{-1}(\mathtt{Th}(T'))$. Similarly the partial order $[T]\sqsubseteq[T']$ can be expressed as $\tau^{-1}(\mathtt{Th}(T))\subseteq\tau^{-1}(\mathtt{Th}(T'))$. Now coming back to our upper-bound candidate $[T^{*}]$ the key is that since ${\cal C}$ is totally ordered we have $\mathtt{Th}(T^{*})=\cup\{\mathtt{Th}(T): [T]\in{\cal C}\}$ and $\tau^{-1}(\mathtt{Th}(T^{*}))=\cup\{\tau^{-1}(\mathtt{Th}(T)):[T]\in{\cal C}\}$. Furthermore $T^{*}$ is consistent. So I have the feeling this works. Please forgive me if I am completely off the mark.

N.B. I should probably have pointed out that if $T\sim T'$ and $T$ is consistent, then $T'$ is also consistent. Otherwise, from $T'\vdash\bot$ we get $T'\vdash\exists! x(\bot)$ and so $T\vdash \exists! x(\bot)$ and finally $T\vdash\bot$.

N.B. I was worried for a second that the above argument would somehow be devoid of substance, by the fact that a maximal element $[T^{*}]$ could simply be obtained by taking the class modulo $\sim$ of a maximal consistent set $T^{*}\supseteq T_{0}$. This does not seem to be the case: $T^{*}$ may be maximal consistent. It does not mean it maximizes the number of mathematical objects it creates, so to speak.