Suppose $G$ and $H$ are groups. $C \subseteq H$ is called a *skew copy* of $G$ in $H$ if $C = hK$ for some $h \in H$ and some subgroup $K$ of $H$ with $K \cong G$.

Question 1:Suppose the infinite symmetric group $S_\mathbb N$ is partitioned into finitely many pieces. Must one of these pieces contain a skew copy of every countable group?

Given two groups $G$ and $H$, let us write $H \rightarrow G$ to mean that whenever $H$ is partitioned into finitely many pieces, one of the pieces contains a skew copy of $G$.

Question 2:If the answer to Question 1 is negative, is it nonetheless true that there is some (possibly very large) group $H$ such that $H \rightarrow G$ for every countable group $G$?

What I know so far about this question is:

$(1)$ The answer to Question 1 is positive if and only if $S_\mathbb N \rightarrow G$ for every countable group $G$.

$(2)$ $S_\mathbb N \rightarrow G$ for every finite group $G$. In fact, something a little stronger is true: for every finite group $G$ and every $r \in \mathbb N$, there is a finite group $H$ such that if $H$ is partitioned into $r$ pieces, then one of them contains a skew copy of $G$.

$(3)$ If $S_\mathbb N$ is partitioned into finitely many Borel pieces (or, a little more liberally, pieces with the Property of Baire), then one of the pieces contains a skew copy of $S_\mathbb N$, and therefore contains a skew copy of every countable group. I do not know whether the Property of Baire can be replaced with Lebesgue measurability.