Let $\Omega$ be a simply-connected domain. Let $z_1,z_2\in\Omega$. Prove that exists an univalent function such that $f(\Omega)=\Omega$ and $f(z_1)=z_2$.
Since $\Omega$ is simply connected, one can apply Riemann mapping theorem to get $g:\Omega\to\mathbb{D}$ ($\mathbb{D}$ is the unit disk) with $g(z_1)=0,g(\Omega)=\mathbb{D}$. I thought taking $$f(z)=g^{-1}(g(z)+g(z_2))$$ thus $f(z_1)=g^{-1}(g(z_1)+g(z_2))=z_2$. On the other hand, $f$ does not map $\Omega$ to itself.
How can I build a map which will satisfy the requirements above?
You're close: by the Riemann mapping theorem there exist biholomorphic $g, h : \Omega \to \mathbb{D}$ such that $g(\Omega) = h(\Omega) = \mathbb{D}$, $g(z_1) = 0$, and $h(z_2) = 0$. Then $f = h^{-1} \circ g : \Omega \to \Omega$ satisfies $f(\Omega) = \Omega$ and $f(z_1) = z_2$.
Edit. As pointed out in the comments, this requires $\Omega \neq \mathbb{C}$. If $\Omega = \mathbb{C}$, a simple affine transformation does the trick.