, 2023

The answer is No. Put $k = \frac{2022}{2023}$ and let $x = 0$ then
$$f(y + a f(y)) = k \cdot \frac{y}{a} + f(0)$$
so $f$ is surjective over $\mathbb{Q}$. Put $x = -a f(y)$, then $f(y) = k \cdot \frac{y}{a} + f(-a f(y))$ it is easy to see that $f$ is injective. So $f$ is bijective. Put $y = 0$, then $f(x + a f(0)) = f(x)$ so $x + a f(0) = x$, resulting in $f(0) = 0$. Replace $x = 0$ then
$$f(y + a f(y)) = k \cdot \frac{y}{a}$$
so $y + a f(y) = f^{-1}(k \cdot \frac{y}{a})$, where $f^{-1}$ is the inverse of $f$. From this, it follows that $y + a f(y)$ is also surjective over $\mathbb{Q}$. Rewrite the problem as
$$f(x + y + a f(y)) = f(y + a f(y)) + f(x)$$
and replace $y + a f(y) = t \in \mathbb{Q}$ then
$$f(x + t) = f(t) + f(x), \forall x, t \in \mathbb{Q}.$$
Therefore, $f$ is additive on $\mathbb{Q}$, thus there exist $c \in \mathbb{Q}$ such that $f(x) = c x, \forall x \in \mathbb{Q}$. Replace to the original equation to get
$$c(x + y + a c y) = k \cdot \frac{y}{a} + c x$$
or
$$\left(c + a c^2 - k \cdot \frac{1}{a}\right) y = 0, \forall y \in \mathbb{Q}.$$
From that we have $(a c)^2 + a c - k = 0$, obviously $\Delta = 1 + 4k$ is not the square of the rational number so the equation of variable $t = a c$ has no rational solution.
Therefore, there does not exist $a \in \mathbb{Q}$ for a function $f : \mathbb{Q} \to \mathbb{Q}$ that satisfies.