The largest group which has embeddings into every nonabelian finite simple group is the Cartesian square of this group.

Let $G$ be a group. The holomorph $Hol(G) = G \rtimes Aut(G)$ can be regarded as a subgroup of the symmetric group $S_{|G|}$, by considering the functions $f: G \to G$ sending $x \in G$ to $x^{\alpha} \cdot g$, for $g \in G$ and $\alpha \in Aut(G)$.
This is never a self-normalizing subgroup of $S_{|G|}$ when $G$ is nonabelian, because any anti-automorphism of $G$, including $x \to x^{-1}$, normalizes it but is not in it. So $N_{ S_{ |G|} } (Hol(G))/Hol(G)$ always has a subgroup of order 2 in this case.

The only nontrivial groups lacking proper subgroups are the cyclic groups of prime order. Among these, only a cyclic group of order 2 can be isomorphic to the unique minimal subgroup of two nonisomorphic finite groups of the same order (a cyclic group and a quaternion group, when the order is a power of 2).

Let $G$ be a nonabelian group in which all subgroups are normal. Then $G$ is isomorphic to the Cartesian product of a abelian torsion group in which all elements have odd order, the quaternion group $Q_{8}$, and a group of exponent 2 (by the Axiom of Choice, this last factor is a vector space over the field $\mathbb{Z}/(2)$, as mentioned before).

Notice how the group of exponent 2 above did not need to be required to be abelian? If $G$ is a group and all cyclic subgroups of $G$ are of order 2, it immediately follows that $G$ is abelian.