Let $G$ be the group of homeomorphisms of unit disk $(D)$ fixing boundary point wise and $P$ be the group of homeomorphisms of plane $\mathbb{R}^2$.
Can we characterize $G$ in terms of $P$. Speaking precisely, can we have the result of the form: $\gamma$ is in $G$ if and only if $F(\gamma)$ is in $P$, where $F$ is some condition or some function (for example it could be in terms of euclidean norm in $\mathbb{R}^2$).
We can easily embedd $G$ in $P$ as: Define a homeomorphism $g:Int(D)\rightarrow\mathbb{R}^2$ in polar coordinates as $g(r,\theta)=(\frac{r}{1-r},\theta)$. Then the function $f:G\rightarrow P$ defined by $f(\gamma)=g^{-1}\gamma g$ is an embedding. Through $f$ we can study the image of $G$ in $P$ rather than $G$ itself.
The only answer is this.
For $h \in P$, let $\phi(h) : D \to \phi(D)$ denote the restriction of $h$ (which is a homeomorphism). Then $\phi(h) \in G$ if and only if $h(x) = x$ for all $x \in S^1 = \partial D$.
The "only if" part is trivial. Conversely, assume $h(x) = x$ for all $x \in S^1$. We know that $S^1$ separates the plane into two connected components $\mathring{D}$ = open unit disk (which is contractible) and $D' = \mathbb{R}^2 \setminus D$ (which is not contractible). $h(\mathring{D})$ must be one of these connected components. But $h(\mathring{D}) \approx \mathring{D}$ is contractible, whereas $h(D') \approx D'$ is not. This shows that $h(\mathring{D}) = \mathring{D}$, i.e. $h(D) = D$. Thus $\phi(h) \in G$.
Edited:
Here are some examples of homeomorphisms of the plane which fix $S^1$ pointwise.
1) Let $f : (0,\infty) \to \mathbb{R}$ be any continuous map such that $f(1) = 0$. Define $$h(z) = \begin{cases} e^{if(\lvert z \rvert)}z & z \ne 0 \\ 0 & z = 0 \end{cases} $$ In the first line we use the complex multiplication on $\mathbb{C} = \mathbb{R}^2$. Note that $\lvert e^{if(\lvert z \rvert)} \rvert = 1$, i.e $h$ maps each circle with radius $r$ around $0$ homeomorphically onto itself.
2) Let $g : [0,\infty) \to [0,\infty)$ be any continuous strictly monotonic increasing function such that $g(0) = 0, g(1) = 1$ and $g(t) \to \infty$ as $t \to \infty$. Define $$h(z) = g(\lvert z \rvert) z .$$