So I have a rather interesting question that came up in some independent research I have been doing on finite groups of small order. I was looking at their (full) subgroup lattices, which included the nodes of said lattice being labelled with the isomorphism class of the subgroups (so I would know the isomorphism class of each subgroup as well as its inclusion relations with other subgroups).
I found out that one can intuit a lot about the automorphism group from studying this lattice, for instance, if a subgroup is the only one of its isomorphism class, then it must be characteristic, and hence is invariant under any automorphism.
Less obvious, is that if a group is not invariant under all automorphisms, then the only possible image subgroups are not only isomorphic, but must have 'similar' principle filters (as described in https://en.wikipedia.org/wiki/Filter_(mathematics)), similar in the sense that they are isomorphic as sublattices, and that the isomorphism classes of the subgroups represented by the nodes coincide. This latter condition is powerful, as it can quickly rule out possible candidate subgroups for the image.
One can also discern certain other consequences from the above result, for instance, in $D_8$ there are two subgroups isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$, the Klein four group, each of which has 3 subgroups isomorphic to $\mathbb{Z}_2$, however, one of those is a common subgroup, namely the center of $D_8$, $Z(D_8)$. Of the remaining 4 distinct subgroups of order 2, they can be 'paired off' by whether or not they are members of the same Klein four subgroup. If an automorphism of $D_8$ would take a non-center order 2 subgroup to another such subgroup with different Klein four intermediate group, then it must do the same to the other order 2 subgroup in the pair (it exchanges the pairs), conversely, if it only moves them within the given pair, then it must also respect the other pair (in general it may or may not act nontrivially on the other pair).
My questions are related to this idea of distinguishability that comes from these observations.
1) Is there a known way of explicitly defining when two subgroups that are isomorphic are distinguishable in the manner described above? Edit: See my answer for this part below.
2) Given a satisfactory answer to 1), it is clear that the above described conditions are necessary, but are they sufficient? (I suspect no, but I think counterexamples will be relatively rare; meaning most automorphisms of most groups will correspond to this idea; part of the problem I'm having is I haven't been able to pin down a reasonably clear definition that fully captures the idea described above).
So I managed to answer 1) of my own question tonight, and finally pinned down some formal definitions for what I have in mind:
As I mentioned in the question, given a subgroup $H$ of a given (finite) group $G$, one can construct the principal filter $\uparrow H$, which is a sublattice of the lattice of subgroups, which has as bottom element, $H$ instead of the trivial group, $G$ remains the top element, and its intermediate elements are exactly the intermediate subgroups, $H<S_i<G$ (I've indexed them so that can refer to distinct such subgroups). For another, distinct subgroup $K$ isomorphic to $H$, likewise we have the principal filter $\uparrow K$ whose intermediate groups we will denote as $K<T_j<G$. If it happens that there exists a lattice automorphism $\alpha$ of the subgroup lattice of $G$, such that the image of $\uparrow H$ is $\uparrow K$, which has the additional property that all subgroups $B$ of $G$ are isomorphic to their images $B\simeq\alpha(B)$. Then we will say that $H$ and $K$ are indistinguishable subgroups of $G$. If there does not exist such a lattice automorphism $\alpha$, then we will say that $H$ and $K$ are fully distinguishable.
There is a strengthening of this condition that can be obtained as follows: Consider an intermediate subgroup $S_i$. If it occurs that this subgroup is a fixed point of the lattice automorphism $\alpha$ (that is, $\alpha(S_i)=S_i$), then $S_i$ is also an intermediate subgroup for $K$, and we have that, as subgroups of $S_i$, $H$ and $K$ are indistinguishable. In this case, we call $H$ and $K$, $S_i$-indistinguishable (as subgroups of $G$). The strongest possible conclusion of this type is that $H$ and $K$ are $S_i$-indistinguishable for every $i\in I$ (that is, for every intermediate subgroup). In this case, we call $H$ and $K$ fully indistinguishable subgroups of $G$. At the opposite extreme, if there are no intermediate subgroups for which $H$ and $K$ are $S_i$-indistinguishable, then we say that $H$ and $K$ are weakly indistinguishable (weakly could be replaced by minimally instead, I suppose).