If $G$ is a domain (connected open set) that is symmetric with respect to the real axis, then $G\cap \mathbb{H} = \{ z\in G : \operatorname{Im} z > 0\}$ is also connected. So if $f\colon G_1 \to G_2$ is a holomorphic map between two domains symmetric with respect to the real axis, then $f(G_1\cap\mathbb{H})$ does not lie entirely above or below the real axis if and only if $f$ attains real values in $G_1\cap \mathbb{H}$. Thus, for the case of a biholomorphic $f$, it is sufficient to have $f(G_1\cap \mathbb{R}) = G_2 \cap \mathbb{R}$ to conclude that $f(G_1\cap \mathbb{H})$ is either the part of $G_2$ in the upper half-plane, or the part of $G_2$ in the lower half-plane.
If there is an $a\in G_1 \cap\mathbb{R}$ with $f(a) \in G_2 \cap \mathbb{R}$ and $f'(a) \in \mathbb{R}$, then $g(z) = \overline{f\left(\overline{z}\right)}$ is also a biholomorphic map $G_1 \to G_2$, with $g(a) = f(a)$ and $g'(a) = f'(a)$. Thus $h = g^{-1}\circ f$ is an automorphism of $G_1$ with fixed point $a$ and $h'(a) = 1$.
We obtain the conclusion if we can show that $h = \operatorname{id}_{G_1}$. For then $g = f$ shows that $f(G_1\cap\mathbb{R}) \subset G_2 \cap \mathbb{R}$, and the same argument for $f^{-1}$ shows the equality.
For that, we don't need any symmetry or other particular properties:
Theorem: Let $G\subset \mathbb{C}$ a domain, and $a\in G$. If $h$ is an automorphism of $G$ with $h(a) = a$ and $h'(a) = 1$, then $h = \operatorname{id}_G$.
Proof: If $G = \mathbb{C}$, then the automorphisms of $G$ are precisely the polynomials of degree $1$, and $h'(a) = 1$ then implies that $h$ is a translation $z \mapsto z+b$, and since $h(a) = a$, it follows that $b = 0$.
If $G = \mathbb{C}\setminus \{p\}$, then $h$ extends to an automorphism $\tilde{h}$ of the Riemann sphere $\widehat{\mathbb{C}}$, and either $\tilde{h}(p) = p,\, \tilde{h}(\infty) =\infty$, in which case $h = \operatorname{id}_{G}$ follows as above, or $\tilde{h}(p) = \infty$ and $\tilde{h}(\infty) = p$, in which case we have
$$\tilde{h}(z) = \frac{c}{z-p}+p$$
for some $c\in\mathbb{C}$. Then $h(a) = a$ yields $c = (a-p)^2$, and we obtain
$$h'(a) = -\frac{(a-p)^2}{(a-p)^2} = -1 \neq 1,$$
so that cannot happen: $h(a) = a$ and $h'(a) = 1$ implies $h = \operatorname{id}_{G}$ also for $G = \mathbb{C}\setminus \{p\}$.
If $\mathbb{C}\setminus G$ contains at least two points, then the family $\operatorname{Hol}(U,G)$ of holomorphic functions on any domain $U$ with values in $G$ is a normal family by Montel's big theorem. In particular - picking $U = G$ - the sequence $\left(h^n\right)_{n\in\mathbb{N}}$ of iterates of $h$ contains a locally uniformly convergent subsequence $\left(h^{n_k}\right)_{k\in\mathbb{N}}$. By Weierstraß' theorem, also the sequences $\left(\frac{d^m}{dz^m}h^{n_k}\right)_{k\in\mathbb{N}}$ of $m$-th derivatives converges locally uniformly for every $m\in\mathbb{N}$. If we already know that $h^{(r)}(a) = 0$ for $2 \leqslant r < m$ (which we vacuously do for $m = 2$), the Taylor expansion of $h$ about $a$ is
$$h(z) = a + (z-a) + c_m (z-a)^m + O\left((z-a)^{m+1}\right).$$
From that we find that the $n$-fold iterate of $h$ has the Taylor expansion
$$h^n(z) = a + (z-a) + n\cdot c_m (z-a)^m + O\left((z-a)^{m+1}\right)$$
by induction:
$$\begin{align}
h^{n+1}(z) &= h\left(a+(h^n(z)-a)\right)\\
&= a + \left(h^n(z)-a\right) + c_m \left(h^n(z)-a\right)^m + O\left(\left(h^n(z)-a\right)^{m+1}\right)\\
&= a + (z-a) + n\cdot c_m(z-a)^m + O\left((z-a)^{m+1}\right)\\
&\qquad + c_m \left((z-a) + O\left((z-a)^m\right)\right)^m + O\left((z-a)^{m+1}\right)\\
&= a + (z-a) + (n+1)c_m(z-a)^m + O\left((z-a)^{m+1}\right).
\end{align}$$
So
$$\frac{d^m}{dz^m}\Biggl\lvert_a h^n = n\cdot \frac{d^m}{dz^m}\Biggl\lvert_a h,$$
and that converges only if $h^{(m)}(a) = 0$. By induction on $m$, it follows that all derivatives of $h$ of order greater than one vanish in $a$, and hence $h = \operatorname{id}_{G}$.
Best Answer
$f_1(z) = 1/z^4$ maps $D$ conformally onto $D_1 = \Bbb C \setminus (-\infty , 1]$, with $f_1(1/2) = 16$.
Then $f_2(z) = \sqrt{z-1}$ maps $D_1$ conformally onto the right halfplane (with the principal value of the square root), with $f_2(16) = \sqrt {15}$.
Finally, $f_3(z) = (z-\sqrt {15})/(z + \sqrt {15})$ maps the right halfplane conformally onto the unit disk, with $f_3(\sqrt {15}) = 0$.