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 = SmallGroup(243,19) and H = SmallGroup(243,20). There is a bijection f:L(G)→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 Aut(G) ≅ 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 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 is induced by a bijection of the underlying sets:
There are no such groups of order dividing 64 (even just having an isomorphism of subgroup lattices respecting normal subgroups).