We know for Brouwer theorem that $f$ (continuous bijective function) have a fixed point. My questions are:
1) Is there a function with only one fixed point $x_0\in Int(\bar{\mathbb{D}}) $ (open disk)?
2) Is there a function with only one fixed point $x_0\in \partial(\bar{\mathbb{D}})$?
For (1) is true since a rotation by any angle $\theta\not=2k\pi$. But for (2) I can't find $f$, I think that for any function with one fixed point on the boundary always have at least on fixed point in the open disk $\mathbb{D}$. Maybe for the continuity of $\partial f$ (the restricion of $f$ to $\partial \mathbb{D}$).
Any ideas or suggestions or maybe a counterexample?
It is useful to know that from the viewpoint of the Riemann sphere, disks and half-planes are the same. The Möbius transformation
$$S \colon z \mapsto \frac{z-i}{z+i}$$
maps the upper half-plane biholomorphically to the unit disk. Now it is easy to see bijections of the closed upper half-plane having exactly one fixed point on the boundary, all translations $T_a \colon z \mapsto z+a$ with $a\in \mathbb{R}\setminus \{0\}$ have $\infty$ as their only fixed point. Conjugating such a translation with $T$ then gives us a homeomorphism (even a holomorphic one) of the closed unit disk with itself having exactly one fixed point on the boundary.
Explicitly, $S^{-1}(z) = i\frac{1+z}{1-z}$, then
$$S\circ T_a \circ S^{-1} \colon z \mapsto \frac{a+(2i-a)z}{2i+a-az}$$
is, for every $a \in \mathbb{R}\setminus \{0\}$, an automorphism of the unit disk with $1$ as its only fixed point.