First, note that by surjectivity, we can find $x_0$ such that $g(x_0) = -f(y_0)$, the problem statement then implies $f(x_0) = -g(y_0)$. By setting $y = y_0$, $x = x_0$ in the problem statement, we have:
$$|g(y) - g(y_0)| = |f(y) - f(y_0)|$$
Note that $f(x)$ is a function of $g(x)$ because if $g(x_0) = g(x_1)$, then according to the above relation, $f(x_0) = f(x_1)$. Knowing the value of $f$ at three points $g(x_0) = 0$, $g(x_1) = 1$, $g(x_2) = i$, the value of $f$ is uniquely determined. Because for each point $x$, there is a unique point $f(x)$ such that the distance of $f$ from
$$f(x_0), f(x_1), f(x_2)$$
would be equal to the distance of $g(x)$ from
$$g(x_0), g(x_1), g(x_2)$$
According to the relation we obtained, the image of
$$f(x_0), f(x_1), f(x_2)$$
must be an isosceles right triangle with side length one, any such triangle can be obtained by a reflection across the x-axis, a translation, and a rotation from the triangle with vertices $(0, 0)$, $(1, 0)$, $(0, 1)$. Therefore, $f(x) = c g(x) + d$ or $f(x) = c \overline{g}(x) + d$ where $c$ is on the unit circle. By substituting the first expression and surjectivity in the original problem statement, we have:
$$|c x + y + d| = |c y + x + d|$$
If $c \neq -1$, then the two expressions $c x + y$ and $c y + x$ can be made equal to any arbitrary two complex numbers, which leads to a contradiction. If $c = 1$, the statement is clearly true. If $c = -1$, then $|z + d| = |-z + d|$ implies $d = 0$. If we substitute the second expression, we have:
$$|c x + y + d| = |x + c \overline{y} + d| = |c x + y + c d|$$
Therefore, since $c x + y$ can be any arbitrary complex number, it is necessary and sufficient that:
$$d = \overline{c} d \implies d^2 = c|d|$$
Therefore, the solutions are:
$$\begin{aligned} f(x) &= g(x) + d \\ f(x) &= -g(x) \\ f(x) &= c^2 g(x) + r c \end{aligned}$$
where $c = 1$ implies $r \in \mathbb{R}$.