Let $A,B,C,D,E,F$ be vector spaces over a field.

Let $x\in A \otimes B \otimes C$ and $y \in D \otimes E \otimes F$ be tensors that are semistable with respect to the natural actions of $\text{SL}(A) \times \text{SL}(B) \times \text{SL}(C)$ and $\text{SL}(D) \times \text{SL}(E) \times \text{SL}(F)$ respectively.

Then $x \otimes y \in A \otimes B \otimes C \otimes D \otimes E \otimes F$ is certainly semistable with respect to the natural action of $\text{SL}(A) \times \text{SL}(B) \times \text{SL}(C)\times \text{SL}(D) \times \text{SL}(E) \times \text{SL}(F)$.

Is $x \otimes y \in A \otimes B \otimes C \otimes D \otimes E \otimes F$ also necessarily semistable with respect to the natural action of $\text{SL}(A \otimes D) \times \text{SL}(B \otimes E) \times \text{SL}(C \otimes F)$?

This seems "too good to be true", especially given how large $\text{SL}(A \otimes D)$ can be relative to $\text{SL}(A) \otimes \text{SL}(D)$, so my current inclination is to believe the answer is no.

On the other hand, the obvious place to look for a counterexample won't work. We might try to use the Hilbert-Mumford criterion to show that $x \otimes y$ is unstable. The obvious one-parameter subgroups of $SL(A \otimes B)$ to try are those acting diagonally on some basis of $A \otimes B$ obtained as a product of some basis of $A$ and some basis of $B$, and similarly for the other groups. If the Hilbert-Mumford criterion for instability is verified for such a one-parameter subgroup, then then $x$ or $y$ is necessarily unstable.

For some very special $x$ and $y$ (e.g. diagonal tensors), one can check that the only possible one-parameter subgroups are of this form, so we can show the answer is yes. In particular, if $\dim A = \dim B = \dim C=2$ then there is only one possibility for $x$, and one can check in this case that the answer is yes. I'm pretty sure I can check it also if the dimensions are 2,2,3 and 2,3,3, but 3,3,3 is already a mystery to me.