(For clarity, I'll use $R,S$ for binary relation symbols and $A,B$ for actual binary relations.)
There is an equality between the numbers (up to isomorphism in the appropriate sense) of partitions of a finite set into at most two pieces and of partitions of that set into pieces with at most two elements. For example, when $n=4$, there are only three "truly different" partitions into at most two pieces, namely $n=4=1+3=2+2$. Similarly there are only three "truly different" partitions into pairs-or-singletons, namely $4 = 1 + 1 + 1 + 1 = 1 + 1 + 2 = 2 + 2$. The fact that these two numbers are always equal has a nice picture proof (rotate the Young diagrams).
Formally, this equality can be implemented "definably" in the following sense:
$(\star)$ There is a first-order $\{R\}$-sentence $\lambda$ and a first-order $\{R,S\}$-formula $\theta$ such that for every finite set $X$ the following hold:
There is at least one $A\subseteq X^2$ with $(X;A)\models\lambda$.
Whenever $(X;A)\models\lambda$ and $B$ is an equivalence relation on $X$ where each class has size $\le 2$, $\theta^{(X;A,B)}$ is an equivalence relation on $X$ with $\le 2$ classes.
If $(X;A)\models\lambda,(X;A')\models\lambda$, and $B,B'$ are partitions of $X$ into pairs-or-singletons then we have $$(X;B)\cong (X;B')\quad\iff\quad(X;\theta^{(X; A,B)})\cong(X;\theta^{(X; A',B')}).$$
Specifically, the following formulas do the job:
$\lambda$ asserts that $R$ is a tournament: for each pair of distinct elements $x,y$, exactly one of $xRy$ or $yRx$ holds.
$\theta(x,y)$ says that either each of $x$ and $y$ is the $R$-greatest elements of their $R$-class, or neither is.
I'm curious whether this construction is optimal with respect to the "auxiliary structure" (the tournament part) involved.
Question: Suppose $\lambda,\theta$ witness $(\star)$ above. Must there be infinitely many $n\in\mathbb{N}$ such that, if $\vert X\vert=n$ and $(X;A)\models\lambda$, $(X;A)$ has a definable-without-parameters tournament?
