Find fixed point domain in $\mathbb{R}^n$ other than $D^n$. A fixed point domain is a subset $F \subset \mathbb{R}^n$ in which Brouwer’s Fixed Point Theorem is true. Can we find fixed point domain in $\mathbb{R}^n$ other than the space homeomorphic to $D^n$? Is there any non-trivial example?
More specifically, can we classify all fixed point domain in $\mathbb{R}$, or even $\mathbb{R}^n$?
 A: I assume this is what you are referring to:

Definition. Let $X$ be a topological space. We will say that $X$ has the Fixed Point Property (FPP) if for any continuous mapping $f:X\to X$ there is $x\in X$ such that $f(x)=x$.

With that definition the Brouwer's Fixed Point Theorem can be reduced to:

Brouwer's Fixed Point Theorem. Closed disk $D^n\subseteq\mathbb{R}^n$ has FPP.

But there are more nonhomeomorphic subspaces of $\mathbb{R}^n$:

Lemma. Let $X$ be a FPP space. If $A\subseteq X$ is a retract of $X$ then $A$ has FPP.

Proof. Take the retraction $r:X\to A$, the inclusion $i:A\to X$ and consider any continuous function $f:A\to A$. Then we have an induced $F=i\circ f\circ r$ and since $F:X\to X$ then $F$ has a fixed point, say $x_0\in X$. We have $x_0=F(x_0)=i(f(r(x_0))$ and since the image of $i$ is $A$, then $x_0\in A$. It can be easily checked that $x_0$ is a fixed point of $f$. $\Box$
Non-trivial Example. Take $D^2$ and perform "squeezing" along its equator towards the center of the ball. The resulting space $X$ is the wedge sum (or bouquet) of two disks $X\simeq D^2\vee D^2$. It is a retract of $D^2$ and thus it has FPP by the lemma. But it is not homeomorphic to any ball because the base point becomes a cut point while disks have no cut points.
The problem of classifying FPP spaces is still open and not much has been done in the field (seems to be very hard). For more information read this: https://en.wikipedia.org/wiki/Fixed-point_property
EDIT. If you are looking for a non-contractible FPP subspace of $\mathbb{R}^n$ then have a look at this:

This space, call it $\mathcal{K}$ can be formally defined as follows:
$$K := \{(x,y)\in\mathbb{R}^2\ |\ \lVert(x,y)\rVert=1\mbox{ and }y\leq 0\}$$
$$\forall_{v,w\in\mathbb{R}^2}\ I(v,w):=\{tv+(1-t)w\ |\ t\in[0,1]\}$$
$$\forall_{n\in\mathbb{N}}\ K_n:= I\big((1, 0), (-1, n^{-1})\big)$$
$$\mathcal{K}=K\cup\bigcup_{n=1}^{\infty}K_n$$
The proof that $\mathcal{K}$ has FPP and it is not contractible can be found in D.R. Smart, "Fixed Point Theorems", Theorem 3.2.3.
