Are there structures that "nontrivially" let us convert ordered pairing functions to unordered pairing functions? This question is an outgrowth of my partial answer to this older MSE question. Throughout, formulas are allowed to include parameters, "definable" means "definable with parameters," all structures are infinite, and all languages are finite.

Kuratowski's definition of an ordered pair in terms of set membership shows more broadly that we can always, in a definable way, turn an unordered pairing function into an ordered pairing function. Going the other way is more difficult; I'm interested in when it is possible.
Given a structure $\mathcal{X}$ in a language $\Sigma$ and a binary function symbol $F\not\in\Sigma$, say that $\mathcal{X}$ is symmetrizable iff there is a $\Sigma\sqcup\{F\}$-formula $\varphi(x,y,z)$ such that whenever $\mathcal{X}'$ is an expansion of $\mathcal{X}$ gotten by interpreting $F$ as an injection $X^2\rightarrow X$, the relation $G:=\varphi^{\mathcal{X}'}$ satisfies the following conditions:

*

*For each $a,b\in\mathcal{X}$ there is exactly one $c\in\mathcal{X}$ such that $G(a,b,c)$ holds. (So $G$ is the graph of a function.)


*For all $a,b,a',b',u\in\mathcal{X}$ we have $G(a,b,u)\wedge G(a',b',u)\implies (a,b)=(a',b')$ or $(a,b)=(b',a')$.


*For all $a,b,c\in\mathcal{X}$ we have $G(a,b,c)\iff G(b,a,c)$.
Basically, $\mathcal{X}$ is symmetrizable iff ordered pairing functions on $\mathcal{X}$ can be definably turned into unordered pairing functions on $\mathcal{X}$. There are two obvious types of symmetrizable structures:

*

*If $\mathcal{X}$ already has a definable unordered pairing function, then $\mathcal{X}$ is trivially symmetrizable.


*A bit more interestingly, if $\mathcal{X}$ has a definable linear order $\triangleleft$ then $\mathcal{X}$ is symmetrizable, via $\varphi(x,y,z)\equiv (x\triangleleft y\wedge F(x,y)=z)\vee(\neg x\triangleleft y\wedge F(y,x)=z)$.
In fact, the second example class doesn't need linear orders as such, but merely tournaments. A tournament on a set $X$ is a function $t:X^2\rightarrow X$ such that $t(a,b)=t(b,a)\in\{a,b\}$ for all $a,b\in X$; if $\mathcal{X}$ has a definable tournament, then $\mathcal{X}$ is symmetrizable.
I'm curious whether this is all that there is (I suspect not!).

Question: Is there a symmetrizable structure with no definable tournament and no definable unordered pairing function?

 A: Let $M = (\{0,1,2\} \times \mathbb{Q}, \triangleleft)$, where $(n,r) \triangleleft (m,s)$ if and only if $r < s$. We'll write $x \equiv y$ to mean $\neg x \triangleleft y \wedge \neg y \triangleleft x$. This is clearly an equivalence relation.
$M$ does not have any tournaments. Given some definable function $f(x,y)$, find an $\equiv$-class $X$ greater than any of the parameters of $f$. The automorphism group of $M$ is transitive on this set, so $f(x,y) = x$ implies $f(y,x)=y$, which is a contradiction.
It is possible to show that $M$ does not have an unordered pairing function, but it is somewhat tedious to write out a complete argument. Essentially you can show that any definable binary function $H(x,y)$ is piecewise equal to either $x$, $y$, or one of $H$'s parameters away from the $\equiv$-diagonal (and it's not too hard to characterize what can happen on the $\equiv$-diagonal for that matter). Furthermore, there must be a 'rectangle' on which one of these cases occurs everywhere. This will clearly always fail to be an unordered pairing function.
Now we'll show that $M$ is symmetrizable. Fix $3$ distinct parameters $a$, $b$, and $c$. Let $F:M^2\to M$ be an injection. I will describe an unordered pairing function $G:M^2 \to M$ which will clearly be $\{a,b,c\}$-definable using $F$. Given $x$ and $y$, consider the set $$A(x,y) = \{F(x,y),F(y,x),F(F(x,y),F(x,y)),F(F(y,x),F(y,x))\}.$$
Note that $A(x,y) = A(y,x)$. Since our $G(x,y)$ will really be a function of $A(x,y)$, we have that $G(x,y) = G(y,x)$.
If $x = y$, let $G(x,y) = F(F(x,x),a) = F(F(y,y),a)$. Assuming that $x\neq y$, note that we cannot have $|A(x,y)|<4$ (because $F(x,y) \neq F(z,z)$ for any $z$, and $F(F(x,y),F(x,y)) \neq F(F(y,x),F(y,x))$).
Let $B(x,y)\subseteq A(x,y)$ be the $\triangleleft$-least $\equiv$-class of $A(x,y)$.

*

*If $B(x,y) = \{F(z,w)\}$ for $z,w \in \{x,y\}$, let $G(x,y) = F(F(z,w),a)$.

*If $B(x,y) = \{F(F(z,w),F(z,w))\}$ for $z,w \in \{x,y\}$, let $G(x,y) = F(F(z,w),a)$.

*If $B(x,y) = \{F(x,y),F(y,x)\}$, let $G(x,y) = F(z,b)$, where $z$ is the third element of the full $\equiv$-class extending $B(x,y)$. (This will exist since $x\neq y$ and $F(x,y)$ is injective.)

*If $B(x,y) = \{F(F(x,y),F(x,y)),F(F(y,x),F(y,x))\}$, let $G(x,y) = F(z,c)$, where $z$ is the third element of the full $\equiv$-class extending $B(x,y)$. (This will exist since $F(x,y) \neq F(y,x)$.)

*If $B(x,y) = \{F(x,y),F(y,x),F(F(z,w),F(z,w))\}$ for $z,w \in \{x,y\}$, let $G(x,y) = F(F(z,w),a)$.

*If $B(x,y) = \{F(z,w),F(F(x,y),F(x,y)),F(F(y,x),F(y,x))\}$ for $z,w \in \{x,y\}$, let $G(x,y) = F(F(z,w),a)$.

Finally note that $B(x,y) = A(x,y)$ cannot happen.
To see that this is in fact an unordered pairing function, we have that $G(x,y)=u$ for some $y$ if and only if

*

*$u = F(z,a)$ and $x$ is one of the $F$-coordinates of $z$,

*$u = F(z,b)$ and $x$ is one of the $F$-coordinates of either of the other two elements of the $\equiv$-class of $z$, or

*$u = F(z,c)$ and $x$ is one of the $F$-coordinates of either of the $F$-coordinates of either of the other two elements of the $\equiv$-class of $z$.

