An equivalence relation on regions of the plane. Let $R\subseteq\mathbb{R}^2$. Consider the set of all "horizontal sections" $H_R =${$Rb|b\in\mathbb{R}$}, where $Rb=${$a\in\mathbb{R} | (a,b)\in R$}. Similarly consider the set of "vertical sections" of $R$, $V_R =${$ aR|a\in\mathbb{R}$} where $aR=${$ b\in\mathbb{R} | (a,b)\in R$}.  Now define the equivalence relation on $\wp (\mathbb{R^2})$ such that $R \sim S$ if, and only if, $H_R=H_S$ and $V_R=V_S$.
QUESTION: What is the equivalence class of a disk?
 A: Denote with $\pi_1,\pi_2: \Bbb R^2 \to \Bbb R$ the coordinate projections.
First, let me make explicit what exactly the elusive sets $H_R$ and $V_R$ are (since I tripped into a pitfall about that earlier on).
We have that $H_R = \{Rb: b \in \Bbb R\}$, where $Rb = \{a \in \Bbb R: (a,b)\in\Bbb R\}$. In terms of the projections, this comes down to $H_R = \{\pi_1[R \cap \pi_2^{-1}(b)]: b \in \Bbb R\}$. We can imagine this as the sets we see when moving a horizontal slit over the set $R$.
Similarly, $V_R = \{\pi_2[R \cap \pi_1^{-1}(a)]: a \in \Bbb R\}$, and it can be imagined as the subsets of $\Bbb R$ obtained from sliding over $R$ with a vertical slit.

For finding the equivalence class, let us reduce to the entirely generic case of the (closed) unit disk $B$.
Then we observe that $H_B = V_B = \{[-a,a]: a \in [0,1]\}$. The apparent symmetry allows us to restrict attention to the upper right quadrant, since $R$ will have to be extended in a unique way, by said symmetry. We will hereafter drop distinction between $R$ and the intersection of $R$ with the upper right quadrant.
The conditions $H_R = H_B$ and $V_R = V_B$ imply that for each $r,s \in [0,1]$, there are unique $a, b$ with $\pi_1[R \cap \pi_2^{-1}(s)] = [0,a]$ and $\pi_2[R \cap\pi_1^{-1}(r)] = [0,b]$. We define mappings $i_H, i_V$ by $i_H(s) =a$ and $i_V(r)=b$.
Now suppose that for given $a$, we find an $s \in [0,1]$ such that $i_H(s) = a$. If for some $s' > s$ we would have that $i_H(s') = a' > a$, then for $i_V(a')$ to be an interval, we would need $(a',s) \in R$ since $s \le s'$, contradicting that $a' > a$ and $i_H(s)=a$.
This contradiction provides us with the information that $s \le s'$ implies $i_H(s) \supseteq i_H(s')$. A similar reasoning can be applied to obtain that $r \le r'$ implies $i_V(r) \supseteq i_V(r')$.
Now what happens when $s < s'$ and $i_H(s) = i_H(s')$? Then $i_V(i_H(s)) = s'$, while for $a' > i_H(s)$ we have necessarily that $i_V(a') < s$; the above analysis hence renders it impossible that there exists an $a''$ such that $s < i_V(a'') < s'$, contradicting our assumptions.
Therefore, both $r \mapsto i_V(r)$ and $s \mapsto i_H(s)$ are strictly decreasing surjections, and by the above argument they are also injective; A cursory remark is that $i_H(i_V(r))=r$ and $i_V(i_H(s)) = s$ for all $r,s$.
In conclusion, $B \sim R$ iff $R$ is the area below a strictly decreasing bijection $f:[0,1]\to[0,1]$, reflected first the horizontal, then the vertical coordinate.

Real analysis tells us that on intervals, "strictly decreasing bijection" is synonymous with "continuous injection mapping endpoints to opposite endpoints". Thus this answer matches Brian M. Scott's comment.
