The ODE system (see below), where $F$ is a given function together the algebraic condition (SYM) imply that $$\boxed{y(t)=k-x(t)} \tag{*}$$ for some $k$ constant.
The result that $y$ is a translation of $x$ reminds me of Lie symmetries of ODEs. The condition (SYM) feels like, smells like, tastes like a symmetry condition.
But symmetry methods involve changing the coordinates of the ODE to canonical coordinates and here we are not doing this? Or are we? My question is: Is there a way of interpreting (*) as arising from some change of variables as in the symmetry analysis of ODEs? If so, how? Sorry if this is all too obvious.
\begin{align} &F(x,x)\,x^\prime(t)-F(x,y)\,y^\prime(t)-1=0 \tag{ODE_1}\\ \\ &F(y,x)\,x^\prime(t)-F(y,y)\,y^\prime(t)-1=0\tag{ODE_2}\\ \\ &F(x(t),x(t))+F(x(t),y(t))=F(y(t),x(t))+F(y(t),y(t)) \tag{SYM}. \end{align}
We also are assuming the ODE system is not under-determined, that is we can be solve it for $x^\prime$ and $y^\prime$. For almost all $x$ and $y$ we require: $$ \det \pmatrix{ F(x,x) & F(x,y) \\ F(x,y) & F(y,y)}\neq 0 $$
Notice also SYM does not hold for any $x,y$ but only for $x(t)$ and $y(t)$ that solve the ODE system.