Retracted.

alexheinis wrote: First we will do the case $g(0)=0, g(1)=1$ and then we will derive the general case from this. For $x\in C$ we have $|f(x)|=|g(x)|$ and $|f(x)+1|=|g(x)+1|$... Are you sure? Putting $y=0$ only gives $|f(x)|=|g(x)+f(0)|$. Also it seems that there are more solutions, e.g. $f(x)=-g(x)$.

Below is my approach Hopefully someone posts a nice solution The functions that satisfy are $f(z)=g(z)$ $f(z)=-g(z)$ $f(z)=g(z)+c$ $f(z)=w\overline{g(z)}$ $f(z)=\frac{c}{\overline{c}}(\overline{g(z)+c})$ for some complex numbers $c$ and $w$ with $c\neq 0$ and $|w|=1$. These functions can be directly verified to work. We want to use the notion of $g^{-1}(x)$. To do this, if multiple values in $\mathbb{C}$ output $x$ chose $g^{-1}(x)$ to output any fixed one of these (chosen arbitrarily). Now let $h(x)=f(g^{-1}(x))$. Letting $x=g^{-1}(a)$ and $y=g^{-1}(b)$ gives $$|h(a)+b|=|a+h(b)|$$Let $P(a,b)$ denote the above assertion. We now split into cases. Case 1: $h(0)\neq 0$ Let $h(0)=c$ and let $z$ be a complex number. Then $P(a,0)$ gives $|h(a)|=|a+c|$. The assertion $P(z,-h(z))$ implies that $h(-h(z))=-z$ so $|-h(z)+c|=|z|$. But we know that $|h(z)|=|z+c|$ [asy][asy] size(4cm, 0); defaultpen(fontsize(8pt)); Label Lc = Label("$c$", position=MidPoint); draw((0,0) -- (3,1), arrow=Arrow(TeXHead), L=Lc); Label Lz = Label("$z$", position=MidPoint); draw((3,1) -- (5,6), arrow=Arrow(TeXHead), L=Lz); draw(circle((0,0),7.811),black+linewidth(.5)+dashed); draw((3,1)--(-6,5), grey+dashed+linewidth(.3)); Label Lz = Label("$h$", position=MidPoint); draw((0,0) -- (-6,5), arrow=Arrow(TeXHead), L=Lz); dot((0,0)); [/asy][/asy] so looking at this geometrically gives that either $h(z)=z+c$ or $h(z)=\frac{c}{\overline{c}}(\overline{z+c})$. (This is because we know that $h(z)$ has to lie on the circle with radius $|c+z|$ and it must be such that $|-h(z)+c|$ or the distance between $h(z)$ and $c$ is $|z|$ which is true for only two points on the circle, $c+z$ and the reflection of $c+z$ over $c$). Notice that if $z$ is a scaler multiple of $c$ then these two are the same so then if there exists $a,b$ not parallel to $c$ and if $h(a)$ follows $h(a)=a+c$ and $h(b)=\frac{c}{\overline{c}}(\overline{b+c})$ then $$|a+c+b|=|a+\frac{c}{\overline{c}}(\overline{b+c})|=|a+c+\frac{c}{\overline{c}}\overline{b}|$$Using the above reasoning, then as $b\neq \frac{c}{\overline{c}}\overline{b}$ we must have that $(a+c)^2$ and $ b\cdot \frac{c}{\overline{c}}\overline{b}$ are parallel or that $(a+c)^2$ and $c^2$ are parallel, but this is clearly impossible. Thus $h$ must follow one or the other. Case 2: $h(0)=0$ Then $P(a,0)$ gives that $|h(a)|=|a|$. The assertion $P(a,h(a))$ gives that $2|h(a)|=|a+h(h(a))|$ so we must have $h(h(a))=a$. Then let $h(1)=w$ for some $|w|=1$. The assertion $P(1,x)$ gives that $|1+x|=|w+h(x)|$. Thus the angles formed by $1,x$ and $w,h(x)$ with the origin must be equal, so either $h(x)=wx$ or $h(x)=w\overline{x}$. But if $h(x)=wx$ then $h(h(x))$ is either $\overline{x}$ or $w^2 x$ so we must have either $w=\pm 1$ or $x$ is real. But if $x$ is real the above expressions are equivalent so we can assume $w=\pm 1$ if the second one does not always hold. But as above if non-real $a$ and $b$ are such that $h(a)=wa$ and $h(b)=w\overline{b}$ then $$|wa+b|^2=|a|+|b|+wa\overline{b}+\overline{wa}b\neq |a|+|b| +w\overline{ab}+\overline{w}ab=|a+w\overline{b}|^2$$so $h$ must be either one or the other leading to the above solutions. Now if $x=g(a)=g(b)$ and we set $g^{-1}(x)=a$ then if instead we let $g^{-1}(x)=b$ then we must have $h(a)=h(b)$ as all listed $h$ differ in more than one point. This suffices to imply that $f(x)=h(g(x))$, finishing.

Let $h(x)=f(g^{-1}(x))$ which implies that $|h(a)+b|=|a+h(b)|$ for each $a,b \in \mathbb{C}$. By $P(a,-h(a))$, $h(-h(a))=-a$ So by $P(a,-h(c))$ , $|h(a)-h(c)| = |a-c|$ for each $a,c \in \mathbb{C}$. So $h$ is an isometry in the plane. But it's well-known that the isometries in the complex plane , are of the form $z \to a + \omega z$ or $z \to a+ \omega \overline{z}$ where $|\omega |=1$so $f(z)=a+\omega g(z)$ for each $z$ , or $f(z) = a + \omega \overline{g(z)}$ for each $z$.The rest is trivial.

Here is a pure algebraic solution. Let $P(x,y)$ be the assertion of the problem. Since $g$ is surjective there exists $a\in \mathbb C$ such that $g(a)=0$. Then $P(x,a)$ implies that there is complex number $C=f(a)$ such that $$|f(x)|=|g(x)+C|.\;\; \;\;\;\;(1)$$By the relation $|z|^2=z\overline{z}$ we can restate $P(x,y)$ as $$|f(x)+g(y)|^2=|f(y)+g(x)|^2 \to (f(x)+g(y))\overline{(f(x)+g(y))}=(f(y)+g(xy))\overline{(f(y)+g(x))}$$$$\to |f(x)|^2+|g(y)|^2+f(x)\overline{g(y)}+g(y)\overline{f(x)}=|f(y)|^2+|g(x)|^2+f(y)\overline{g(x)}+g(x)\overline{f(y)}.$$Define the function $h(x)$ as $h(x):= f(x)-C$. Then by (1) we have $$h(x)\overline{g(y)}+\overline{h(x)}g(y)=h(y)\overline{g(x)}+\overline{h(y)}g(x).\;\;\;\; (2)$$$g$ being surjective implies there are $e,d \in \mathbb C$ such that $g(e)=1$ and $g(d)=i$. Therefore, by taking $y$ equal to $e,d$ in (2) we deduce that $$2\cdot \mathrm{Re}\left(h(x)\right) = z_1g(x)+\overline{z_1 g(x)},\;\;\; 2\cdot \mathrm{Im}\left(h(x)\right) = z_2g(x)+\overline{z_2 g(x)}$$where $z_1= \overline{h(e)}$ and $z_2 = \overline{h(d)}$. This implies that $$2\cdot h(x)= \left[z_1g(x)+\overline{z_1 g(x)}\right]+ i\left[ z_2g(x)+\overline{z_2 g(x)}\right].$$Consequently we infer that there are complex numbers $A,B,C$ such that for all $x\in \mathbb C$ we have that $$\boxed{f(x)=Ag(x)+B\overline{g(x)}+C}$$In the rest we need to find such $A,B,C$ such that the assertion $P(x,y)$ follows. We can write this assertion as $$|Ag(x)+B\overline{g(x)}+g(y)+C|=|Ag(y)+B\overline{g(y)}+g(x)+C|\;\; \text{for all}\; x,y\in \mathbb C.$$Since $g$ is surjective then this relation is equivalent to $$|Ax+B\overline x+y+C|=|Ay+B\overline y+x+C|\;\; \text{for all}\; x,y\in \mathbb C.$$The rest follows as a fun exercise to $$\boxed{f(x)=g(x)+s},\;\;\boxed{f(x)=-g(x)},\;\; \boxed{f(x)=b^2\overline{g(x)}+rb}$$where $|b|=1,r\in \mathbb R$ and $s\in \mathbb C$.

I was scared about guessing the answer of this problem, but turns out surjective is strong and idea is straight.