Does the order, lattice of subgroups, and lattice of factor groups, uniquely determine a group up to isomorphism?

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: