Subsets of $G$-sets with sharply transitive $G$-action

Question

Let $G$ be an infinite group acting sharply transitively on a set $X$. Let $Y\subset X$ be a proper subset.

1. Is there a subgroup $H\leq G$ which acts sharply transitively on $Y$ ?

I think this is not true but I couldn't come up with a counterexample immeditely.

2. For which $Y\subset X$ such a subgroup $H$ exists?

Answer 1

2) Clearly any $Y$ that has such a $H$ must be equal to $H\cdot Y$, and by transitivity all such $Y$ must be of the form $H\cdot x$ for at least one $x \in Y$.

Now clearly for any $x \in X$ and any $H \leq G$, the orbit $H\cdot x$ has the property that $H\cdot (H\cdot x) = (HH)\cdot x = H\cdot x$, so $H$ acts transitively on $H\cdot x$, and freeness is inherited by $G$.

Therefore the $Y$ for which this property holds are exactly the orbits of points by subgroups of $G$.

1) Pick any $Y \subset X$ that is not an orbit of a point under a subgroup of $G$, for example (since $X$ is infinite) any cofinite $Y \subsetneq X$.