No, the lattice of subgroups, the lattice of normal subgroups, the order of the group, and the automorphism group do not (even taken together) determine the isomorphism type of a finite group.
Take $G = \mathrm{SmallGroup}(243, 19)$ and $H = \mathrm{SmallGroup}(243, 20)$. There is a bijection $f \colon L(G) \to L(H)$ between their lattices of subgroups such that:
- $|X| = |f(X)|$,
- $X ≅ f(X)$ unless $X = G$,
- $X ≤ Y$ iff $f(X) ≤ f(Y)$,
- $X ⊴ G$ iff $f(X) ⊴ f(G) = H$,
- $G/X ≅ H/f(X)$ whenever $X ≠ 1$ is normal.
Additionally, $\operatorname{Aut}(G) ≅ \operatorname{Aut}(H)$. The fourth bullet shows in particular, that $f$ induces an isomorphism between the lattice of quotient groups of $G$ and the lattice of quotient groups of $H$. The second and fifth bullets show the isomorphism respects everything about the subgroups’ properties as abstract groups.
The groups $G$ and $H$ have presentations
\begin{align*}
G &= \bigl\langle a, b, c \mid a^{27} = b^3 = c^3 = 1,\ ba = abc,\ ca = acz,\ cb = bcz \bigr\rangle\text{ where $z = a^9$} \,, \\
H &= \bigl\langle a, b, c \mid a^{27} = b^3 = c^3 = 1,\ ba = abc,\ ca = acz,\ cb = bcz \bigr\rangle\text{ where $z = a^{-9}$} \,.
\end{align*}
The function $f$ is induced by a bijection of the underlying sets:
- $f(a^i b^j c^k) = a^i b^j c^k$.
There are no such groups of order dividing $64$ (even just having an isomorphism of subgroup lattices respecting normal subgroups).
