In set theory "equiconsistency" (and not "consistency") of the theories is the main part of researches. So we usually try to construct a new model using a given one. In the other words we "transform" a given model to another one by some kind of "operations". Even it seems that we can do this transformation in the realm of syntax instead of semantics. For example if $T$ be a theory which is produced from $ZFC$ by replacing any phrase $x\in y$ with $\neg (x \in y)$ then $T$ is equiconsistent with $ZFC$ because for all set (or proper class) $M$ and for all binary relation $E$ on it we have:
$\langle M,E \rangle \models ZFC \Longleftrightarrow \langle M,M\times M\setminus E \rangle \models T$
Now define:
Definition (1): Let $\mathcal{L}$ be a first order language. A function $f:\mathcal{L}-Form\longrightarrow \mathcal{L}-Form$ called a "syntactical transformation" if it preserves logical operators, i.e.
(1)$\forall n\in \omega~~~\forall \varphi (x_{1},...,x_{n})\in \mathcal{L}-Form~~~~~$
The number of free variables of the formula $f(\varphi (x_{1},...,x_{n}))$ is equal to
the number of free variables of the formula $\varphi (x_{1},...,x_{n})$
(2) $\forall n\in \omega~~~\forall \varphi (x_{1},...,x_{n})\in \mathcal{L}-Form$
$f(\neg \varphi (x_{1},...,x_{n}))=\neg f(\varphi (x_{1},...,x_{n}))$
(3) $\forall m,n\in \omega~~~\forall \varphi (x_{1},...,x_{m}),\psi (y_{1},...,y_{n})\in \mathcal{L}-Form$
$f(\varphi (x_{1},...,x_{m})\wedge \psi (y_{1},...,y_{n}))=$
$f(\varphi (x_{1},...,x_{m}))\wedge f(\psi (y_{1},...,y_{n}))$
(4) $\forall n\in \omega~~~\forall \varphi (x, y_{1},...,y_{n})\in \mathcal{L}-Form$
$f(\exists x~\varphi (x,y_{1},...,y_{n}))=\exists x~f(\varphi (x, y_{1},...,y_{n}))$
Remark (1): In some sense a syntactical transformation is a "homomorphism" from $\mathcal{L}-Form$ to itself when we look at the set $\mathcal{L}-Form$ as a structure endowed with unary and binary logical operators $\neg~,~\exists~,~\wedge$.
Remark (2): If $f$ be a syntactical transformation in the first order language $\mathcal{L}$, then the values of $f$ on $\mathcal{L}-Form$ are determined by the values of $f$ on the atomic $\mathcal{L}$-formulas.
Remark (3): By the condition (1) in the definition (1) it is easy to see that for any syntactical transformation $f$ in a first order language $\mathcal{L}$ and for any $\mathcal{L}$ -sentence $\sigma$, $f(\sigma)$ is a sentence too. So for any $\mathcal{L}$-theory $T$ the image of $T$ under the function $f$ ($f[T]=\lbrace f(\sigma)~|~\sigma \in T\rbrace$) is an $\mathcal{L}$-theory too.
Definition (2): A syntactical transformation $f$ in the language $\mathcal{L}$ called "consistency preserving" if:
$\forall T\in \mathcal{L}-Theory~~~Con(T)\Longrightarrow Con(f[T])$
Remark (4): The identity map is a trivial consistency preserving syntactical transformation in any first order language.
Question (1): Is there a non-trivial consistency preserving syntactical transformation in any first order language? Is there such function for the language of set theory $\mathcal{L}=\lbrace \in \rbrace$?
Definition (3): Assume that $T$ is a theory in the first order language $\mathcal{L}$. Now define the set $Trans(T)$ to be as follows:
$\lbrace f~|~f~\text{is a syntactical transformation and}~f[T]~\text{is a consistent theory} \rbrace$
Remark (5): For example in the language of set theory the mapping which is given by $x=y\mapsto x=y$ and $x\in y \mapsto \neg (x\in y)$ is in $Trans(ZFC)$. But the map which is given by $x=y\mapsto \neg (x=y)$ and $x\in y \mapsto x= y$ is not in $Trans(ZFC)$ because it transforms the Axiom of Extensionality into a non-satisfiable sentence as follows:
$[\forall x~\forall y~(x=y\leftrightarrow \forall z~(z\in x\leftrightarrow z\in y))]\mapsto [\forall x~\forall y~(\neg (x=y)\leftrightarrow \forall z~(z=x\leftrightarrow z=y))]$
So there is a natural question here:
Question (2): What are the members of $Trans(ZFC)$?
In the other words how free can be the choice of two formulas $\varphi (x,y)$ and $\psi (x,y)$ such that for the syntactical transformation $f$ which is defined by $x=y\mapsto \varphi (x,y)$ and $x\in y\mapsto \psi (x,y)$ we have $Con(ZFC)\Longrightarrow Con(f[ZFC])$?