Let $K \subseteq G$. Nec. and suff. cond. for that $\exists$ a normal subgr $H$ of $G$ s.t $exists$ transversal $I$ s.t $I \subseteq K$.

Question

I am trying to solve/understand the following question that I have come up with;

Let $G$ be a group, and $K \subseteq G$ be given. What are the necessary and sufficient condition for that there exists a normal subgroup $H$ of $G$ such that we can find a transversal $I$ of $G/H$ with with $I = K$.

, or alternatively a more weak result that (as a starting point),

Let $G$ be a group, and $K \subseteq G$ be given. Can we find an equivalence relation on $G$ such that there exists a transversal I of $G/K$ with $I \subseteq K$ ?

I have produced a couple of (really) trivial necessary conditions for the order of $H$ (if exists); however, I do not have any really progress in general, so what can we do ? Can you give me some starting point for even special type of groups ? is there any similar general/special result similar to this (as in here)?