I want to split an Euclidean disk (it should make no difference whether it's open or closed) into two non-intersecting sets of points which are identical in a sense that one set can be transformed into another (and vice versa) by using only shift and rotation.
Can that be done?
No.
Let $D$ be the (open, closed or whatever) unit disk and let $T\colon \mathbb R^2\to\mathbb R^2$ be a rotation or trnslation such that $D=A\cup T[A]$ for some set $A$ with $A\cap T[A]=\emptyset$. If $T[\partial D]=\partial D$, the point $0$ must be a fixed point of $T$; but then $0\in A$ implies $0=T(0)\in B$ and vice versa, contradiction. Hence $T[\partial D]\ne \partial D$, i.e. these two circles intersect in two distinct points $a,b$. Let $C_a$, $C_b$ be the circles through $a,b$ around the center of rotation (whch is on the line $ab$) - or in case of a translation the lines orthogonal to $ab$ through $a, b$, respectively. These circles/lines intersect (not touch) $\partial D$, hence chop the interior of $D$ into three parts with nonempty interior. For the two "outer" parts we have that their images under $T$ are disjoint from $D$, hence thy must be $\subseteq B$. But they are also disjoint from $T[D]$ hence must be $\subseteq A$, contradiction.
This is not possible for closed disks. Say it is possible and you can divide the disk into two parts which are non-intersecting(say $A$ and $B$) and one can be transformed into another (and vice versa) by using only shift and rotation . Use the transformations $$g_1:A \rightarrow B $$ $$g_2:B \rightarrow A $$ Extend them to $\Bbb R^2$ call the extensions $g_1'$ and $ g_2'$.This can be done because they are given to be composition shifts and rotations. Now $g_1'og_2'$ and $g_2'og_1'$ are both $identity$. Hence $g_1og_2$ and $g_2og_1$ are both $identity$. The extension of $g_1$ to the disk will have no fixed point. This is a contradiction to brouwer's fixed point thoerem.
For any other disk use a similar proof on the closure of the disk.
Here is a(n over-complicated) proof which does not appeal the classification of plane isometries.
Proposition: The unit ball in $\mathbb{R}^2$ cannot be partitioned into two disjoint congruent subsets.
Proof: Let $D$ be the (open, closed, whatever) unit disk, and suppose there is a partition $D = X \coprod Y$ into two pieces $X$ and $Y$ that are congruent by an isometry $f: X \to Y$. Denoting the center of the disk by $p \in D$, assume without loss of generality that $p \in X$.
Since $f(p) \not\in X$, we know $f(p) \neq p$, hence $r:= d(p, f(p))> 0$.
Claim: For any $\epsilon \in (0,2)$, there are points $x$, $y\in X$ with $d(x,y) \ge 2-\epsilon$.
Proof: Assume not. Then $m:= \text{diam}(X) = \text{diam}(Y) < 2$. Denote by $A(r,s)$ the open annulus centered at $p$ with inner radius $r$ and outer radius $s$.
Let $x\in A(m/2, 1) \cap X$. Since $d(x,-x) > m$, it must be the case that some small neighborhood $N$ of $-x$ is contained in $Y$. However, for any $y \in N$, the same argument implies $-y \in X$; hence $-N \subset X$. Thus, $X \cap A(m/2, 1]$ is open in $A(m/2, 1)$.
Similarly, it's true that $Y \cap A(m/2,1)$ is open in $A(m/2,1)$. But this gives the connected set $A(m/2,1)$ a decomposition $$A(m/2,1) = (X \cap A(m/2,1)) \cup (Y \cap A(m/2, 1))$$ into disjoint open subsets. $\blacksquare$
One can easily compute that $$\text{diam}(D \cap f(D)) = \sqrt{4-r^2} < 2.$$ From the claim above, there are points $x, y \in X$ such that $d(f(x), f(y)) = d(x,y) > \sqrt{4-r^2}$. But $f(x), f(y) \in D \cap f(D)$, contradiction. $\blacksquare$
