Here are some observations:

$A$ is just a tensor algebra $T(V)$ for $V=K^r=span\{x_1,\ldots,x_r\}=J/J^2$. Let's work with that because choosing bases is evil.

Every morphism $\phi: A/I \to A/I'$ lifts to a (in fact multiple) morphism(s) $\Phi: A\to A$ which maps $I$ into $I'$ simply because polynomial algebras are free algebras. But it is not clear that $\Phi$ needs to be an isomorphism as well and I think it might not be true in general.

But! It would also be nice to have a lift in between so to speak which is an automorphism of $A/J^s$ instead and that works out nicely because everything is local and finite-dimensional.

Lemma 1: Every isomorphism $\phi: A/I\to A/I'$ lifts to an automorphism $\widehat{\phi}$ of $A/J^s$ which sends $I$ to $I'$.

Proof: You require $I$ and $I'$ to be contained in $J^2$, the radical of $A/I$ is $J/I$, every isomorphism of algebras sends the radical to the radical and induces an automorphism of $A/J^2=K\oplus V$. Therefore $\forall x\in J: \Phi(x)\equiv \alpha(x) \mod J^2$ for some $\alpha\in GL(V)$. In particular $\Phi(J)\subseteq J$, thus $\Phi(J^i)\subseteq J^i$ for all $i$ and we get a well-defined endomorphism $\widehat{\phi}$ of $A/J^s$.

Also $im(\widehat{\phi})+J^2/J^s=A/J^s$ so that $\widehat{\phi}$ is surjective by Nakayama's lemma and therefore bijective as a map $A/J^s\to A/J^s$. QED.

Alright, we now face the problem of describing and distinguishing the orbits of $\operatorname{Aut}(A/J^s) \curvearrowright \operatorname{Ideals}(A/J^s)$.

Since automorphisms respects the radical, we have a natural filtration here which gets respected by all automorphisms. This first distinguising invariant of ideals we should consider is where in the filtration $A \supseteq J \supseteq J^2 \supseteq \ldots$ we first see $I$, i.e. what is the smallest degree of a monomial occuring in a generating set $I$, i.e. the smallest number $d(I)\in\mathbb{N}$ such that $I\not\subseteq J^{d+1}$. A second invariant is the subspace $I+J^{d+1} \leq J^d/J^{d+1}$, more specifically its orbit under the $Aut(A/J^s)$-action on $J^d/J^{d+1}$.

Contained in your problem is the classification of all ideals with $d(I)=s-1$, i.e. those $I$ that are sandwiched betweens $J^s$ and $J^{s-1}$. Since $Aut(A/J^s)$ acts on $J^{s-1}/J^s$ like $GL(V)$ on $V^{\otimes(s-1)}$, we find ourselves with

Problem A: For all $s\in\mathbb{N}_{\geq 2}$ classify the orbits of $GL(V)$ on the subspaces of $V^{\otimes s}$. Ideally, find something like a normal form algorithm.

This alone seems incredibly hard. It smells like there is a lot of non-trivial algebraic geometry involved here. I may, computationally it is easy to program in GAP, because one can write down everything completely explicit.

The number of such orbits probably does not depend on $q$ at all, but is probably polynomial in $r$ with degree depending on $s$.

Even just the one-dimensional case is non-trivial, even the case of pure tensors. Since $P(V)\times\cdots\times P(V) \to P(V\otimes\cdots\otimes V)$ is injective, the $GL(V)$-orbits on pure tensors in $V^{\otimes s}$ are the same as the $GL(V)$-orbits on $P(V)^s$ which already gives you a very crude, but very disheartening lower bound for the number of isomorphism classes of ideals:

Corollary: There are at least $2^{\lfloor\frac{s-1}{2}\rfloor}$ isomorphism classes of one-dimensional ideals of $A/J^s$.

Proof: $GL(V)$ has two orbits on $P(V)\times P(V)$ (all pairs of linearly dependent vectors and all pairs of linearly independent vectors respectively). For any $G$-sets $X,Y$, the number of $G$-orbits on $X\times Y$ is at least the number of $G$-orbits on $X$ times the number of $G$-orbits on $Y$ so that $GL(V)$ has $\geq 2^{s/2}$ orbits on $(P(V)\times P(V))^{s/2}$ so that there are at least $2^{\lfloor\frac{s-1}{2} \rfloor}$ many one-dimensional ideals generated by pure tensors in $V^{\otimes(s-1)}$. QED.

**EDIT 03.07.**: Speaking of algebraic geometry: The Grassmannian $Gr_k(V^{\otimes s})$ is a $k(r^s-k)$-dimensional smooth projective variety while $GL(V)$ is $r^2$-dimensional. The orbits can therefore not be more than $r^2$-dimensional. The orbit space itself isn't a scheme itself, I think, but maybe it's a stack or something. Nevertheless my heurestic is that we should expect it to be $\geq (k(r^s-k) - r^2)$-dimensional at least at some points if $s>2$. In particular, there will be something like $q^{\frac{1}{3}r^{3s}+\Theta(r^{2s})}$ isomorphism classes (at least!) of subspaces. And that's even **worse** than exponentially w.r.t. $s$. It is so much worse that we are now in the vicinity of the number of elements of $A$ itself.

Even in your extremely limited set of cases, this works out the be a massive number! If $(q,r,s)=(5,3,4)$, then $q^{\frac{1}{3}r^s}=5^{3^3}\approx 7.4\cdot 10^{18}$. There is no way you can enumerate all of that.

Anyway... Assuming for the moment that somehow we find a way to solve Problem A. In our situation we can decide whether the two invariants $d(I)$ and $d(I')$ as well as the subspaces $I+J^{d(I)+1}$ and $I'+J^{d(I')+1}$ are the same. Ideally we solve problem A in such a way that if they are the same, we get an explicit $\alpha\in GL(V)$ which actually maps these two subspaces onto on another so that we can simplify the problem a bit as we move on to more complex invariants.

More generally we can use a solution to Problem A to distinguish the graded structure which comes with any admissible ideal $I$: We have a filtration of $I$:

$I \supseteq I \supseteq I \supseteq I\cap J^3 \supseteq I \cap J^4 \supseteq ...$

which gives us a graded ideal $gr(I) = \bigoplus_{k=2}^{s-1} I\cap J^k/I\cap J^{k+1}$ in $gr(A/J^s) = \bigoplus_{k=0}^{s-1} J^k/J^{k+1}$. Note that $Aut(A/J^s)$ respects the radical filtration and therefore induces automorphisms of the graded algebra. In fact the graded algebra is just the tensor algebra $T(V)$ (up to degree s) once again, but now $Aut(A/J^s)$ acts as $GL(V)$ on it so that we can use Problem A to use this invariant.

This gives a whole bunch of invariants more.

belowby some double-exponential of the form $q^{O(r^{3s})}$ (see the newest edit to my answer). Even for your smallish numbers, this will make it impossible to actually enumerateallisomorphism classes. The only way out is probably to use a coarser equivalence relation than isomorphism. $\endgroup$ – Johannes Hahn Jul 3 '18 at 16:00