The "contrary" of an isomorphism Roughly, my question is: is there a standard name for functions which one might characterise as the "contrary" of an isomorphism?
Here is a more precise version of my question. Working model-theoretically, consider the following definition.

Definition. Let $\mathscr{L}$ be a relational signature, and let $\mathcal{A}, \mathcal{B}$ be $\mathscr{L}$-structures. An $\mathscr{L}$-anti-embedding $\pi : A \longrightarrow B$ is any injection such that:
$(a_1, \ldots, a_n) \in R^\mathcal{A}$ iff $(\pi(a_1), \ldots, \pi(a_n)) \notin R^\mathcal{B}$, for each $n$-place predicate $R \in \mathscr{L}$ and all $a_1, \ldots, a_n \in A$
A bijective $\mathscr{L}$-anti-embedding is an $\mathscr{L}$-anti-isomorphism. An $\mathscr{L}$-anti-isomorphism from a structure to itself is an $\mathscr{L}$-anti-automorphism.

My question is: are there standard names for the kinds of functions I just defined? Indeed, are they discussed anywhere?
I have only encountered one such function "in the wild". I was reading Thomas Forster on Church-Oswald set theory; he called his $\{\in\}$-anti-automorphism an "antimorphism". (NB that, in the Church-Oswald setting, self-membered sets are perfectly ok.)
Update: Googling the phrase "anti-isomorphism" teaches me that, in group theory, an anti-isomorphism is standardly defined as a bijection $\pi : \mathcal{G} \longrightarrow \mathcal{H}$ such that $\pi(x \cdot^\mathcal{G} y) = \pi(y) \cdot^\mathcal{H} \pi(x)$. That's obviously a totally different idea. So I should use a different name! If there is no standard name, I welcome suggestions!
 A: Tim has drawn my attention to this (Thank you, Tim!)  Perhaps i should provide what Old Lags do in this kind of setting, namely provide some ancient history.
The context is Quine's NF.  Let us say the dual of a formula of the language of set theory is the result of replacing all occurrences of $\in$ by `$\not\in$'. It's routine to show that the dual of any axiom of NF is a theorem of NF.  Thus if we take $<$M,$\in>$ a model of NF and consider M equipped with the complement (in MxM) of the membership relation we get another model of NF!  Are these two structures elementarily equivalent?  Not reliably!  Might they be isomorphic?  If they are, then the isomorphism is an antimorphism.  Of course - since this is NF - there is the possibility that (not only might there be an antimorphism but) the antimorphism is a set of the model!  It's unknown if this can happen.  The best i can do is show that every model of NF is elementarily equivalent (with respect to stratifiable formul{\ae}) to one with two (set) permutations $\sigma$ and $\tau$ satisfying
both
$$(\forall x y)(x \in y \leftrightarrow \sigma(x) \not\in \tau(y))$$ 
and
$$(\forall x y)(x \in y \leftrightarrow \tau(x) \not\in \sigma(y))$$
Arranging for $\sigma = \tau$ is beyond me at the moment. The word
`antimorphism' is a coinage of your humble correspondent, tho' i can't give you chapter and verse.
Sorry if some readers find this perhaps off-piste, but this is the context of Tim's question, and might be helpful.
Thomas Forster www.dpmms.cam.ac.uk/~tf
A: In the case of an anti-isomorphism we could identify $B$ with $A$ in which case $f$ is the identity function but each predicate for $B$ is the negation of that predicate in $A.$  This is essentially what you did in a comment with $(A \times A) \setminus\mathord{\in^\mathcal{A}}.$
For the unary case of each predicate indicating belonging or not belonging to a certain subset,  one gets the set complements.  If the model was a simple graph with loops allowed then we have the complementary graph. There the set is the vertices. If one wished to avoid loops then the set could be the pairs of vertices though that reduces to the previous example. Similarly a linear order and the reverse order. Again the pairs $(a,a)$ need care so either take the set to be ordered pairs of distinct elements or else semi-strong linear orders meaning no assumption of reflexive or non-reflexive.
What does Church-Oswald set theory does with the possibility that $a \in a?$ I suppose that again it could be permitted on an element by element basis. 
