This is covered in Chapter 8 of Schmidt's book "Subgroup lattices of groups"; everything in this answer is there, but I summarize it here for convenience.
Say a group $G$ has a dual $\hat{G}$ if their subgroup lattices are anti-isomorphic. Your question is about when $G$ is dual to itself.
Baer (1939) proved that every group w/ dual is torsion, i.e. all its elements have finite order.
For locally finite groups (which means every finitely generated subgroup is finite), there is a complete classification of groups with duals. (This is Theorem 8.2.2 in Schmidt, due to Suzuki and Zacher):
If $G$ is a locally finite group, and $\delta$ is a lattice anti-isomorphism to $L(\hat{G})$, then:
- $G = \prod_{\lambda} G_{\lambda}$ where the $G_{\lambda}$ are coprime to one another
- $\hat{G} = \prod_{\lambda} \delta(\hat{G}_\lambda)$, where $\hat{G}_\lambda := \langle G_{\mu} : \mu \neq \lambda \rangle$ and the $\delta(\hat{G}_\lambda)$ are coprime to one another
- For every $\lambda$, there exists (not nec. distinct) primes $p,q$ such that one of the following holds:
- $G_\lambda$ is a cyclic $p$-group, $\delta(\hat{G}_\lambda)$ is a cyclic $q$-group, or
- $G_\lambda$ and $\delta(\hat{G}_\lambda)$ are both either elementary abelian of order $p^n$ or a semi-direct product of an elementary abelian group of order $p^{n-1}$ by a group of order $q$, or
- $G_\lambda$ and $\delta(\hat{G}_\lambda)$ are finite non-Hamiltonian $p$-groups with modular subgroup lattices
Maybe from these more can be said when $G$ is self-dual ($G = \hat{G}$)...