As already explained by others, a functional equation between the CDF of two random variables $X$ and $Y$, such as the equation in this question, implies no almost sure relation whatsoever between $X$ and $Y$. A first reason is that $X$ and $Y$ may be defined on different probability spaces. A second reason is that, even if $X$ and $Y$ are defined on the same probability space, their CDF do not determine $X$ and $Y$, at all.
Basic examples to keep in mind in this setting might be $X$ uniform on $(0,1)$ and $Y=1-X$, or $X$ uniform on $(0,1)$ and $Y$ uniform on $(0,1)$ but independent on $X$. Both these $X$ and $Y$, at the level of their CDF, are indistinguishable from $X$ uniform on $(0,1)$ and $Y=X$.
However, a similar, but different, question might be of interest: consider a CDF $G$ and a function $g$ such that $g\circ G$ is also a CDF (hence $G$ replaces $F_X$ and $g\circ G$ replaces $F_Y$). Since every function which is nondecreasing and continuous on the right with limits $0$ and $1$ at $-\infty$ and $+\infty$ is a CDF, one way to ensure that $g\circ G$ is a CDF is to ask that $g$ is nondecreasing, continuous on the right, with limits $0$ and $1$ at $0$ and $1$. Assume that all this holds and that $X$ has CDF $G$, one can ask if there is a way to build a random variable with CDF $g\circ G$, using only $X$. The problem becomes:
Let $X$ with CDF $G$. Find a function $a$ such that the CDF of $a(X)$ is $g\circ G$.
Here is a fact:
If $U$ is uniformly distributed on $(0,1)$ and $G$ is a CDF, there exists a function, which we denote $G^\ast$, such that the CDF of $G^\ast(U)$ is $G$.
The function $G^\ast$ is a pseudo-inverse of $G$, defined as follows: for every nondecreasing and continuous on the right function $h$ and every real number $u$, $ h^\ast(u)=\inf\{x\mid G(x)\ge u\}. $ It is a nice analysis exercise to check that $h^\ast$ may also be defined by the equivalence $ h^\ast(u)\le x\iff u\le h(x). $ Consequences are that $h^\ast$ is nondecreasing and continuous on the right and that $(h_1\circ h_2)^\ast=h_2^\ast\circ h_1^\ast$. This answers a slightly easier question than ours:
Choose $U$ uniformly distributed on $(0,1)$ and consider the random variables $X=G^\ast(U)$ and $Y=(G^\ast\circ g^\ast)(U)$. Then the CDF of $X$ is $G$ and the CDF of $Y$ is $g\circ G$.
This also suggests that, to get $Y$ as a function of $X$, one could consider:
$Y=a(X)$ with $a=G^\ast\circ g^\ast\circ G.$
With this definition, one sees that $ [Y\le y]=[g^\ast(G(X))\le G(y)]=[G(X)\le g(G(y))], $ hence the proof would be complete if one knew that $\mathrm P(G(X)\le z)=z$ for every $z$ in the image of $g\circ G$ (note that to ask this for every $z$ is hopeless in general).
When $g$ is the identity, the problem itself becomes trivial (choose $Y=X$) but our function $a=G^\ast\circ G$ is not always the identity. In general, $G^\ast\circ G(x)\le x$ for every $x$ and it seems that $ (G^\ast\circ G)(x)=\inf\{z\le x\mid\mathrm P(z which would imply that $(G^\ast\circ G)(X)=X$ almost surely.