What i mean by "same iterated set of nullities", is that $$\dim(N(A))=\dim(N(B))\\ \dim(N(A^2))=\dim(N(B^2))\\ \vdots\\ \dim(N(A^{k-1}))=\dim(N(B^{k-1}))\\ N(A^k)=N(B^k)=0$$
My first question is whether this even is a property of matrices. I am led to believe it is, since it seems to be the reason that the block-subdiagonal matrix is in its specific similarity class, and that we can create it by having as many blocks as there are strings in the string basis, and the size of each block being the length of the string it represents. I can, however, not find anything about this potential property of matrices online.
Now for my attempt at "proving", this: one way that we can look at similarity between two matrices $D$, and $C$, is if for any map-basis combination $(t_1, B_1)$ giving $D$, so $Rep_{B_1,B_1}(t_1)=D$, there exists a corresponding map-basis combination $(t_1, B_2)$ giving $C$. One could say that they should have the same range of maps they can represent. If any map $t$ has a nullity of $l,$ so must any representation $Rep_{B,B}(t)$ have that same nullity, because it vaguely said cannot add or remove extra information just by using different bases. Other way around too, if $N(D)=l$, then any map $D$ can represent must have that same nullity. Letting $A,B$ be nilpotent matrices with the same set of nullities, they must both have only an eigenvalue of $0$. Therefore, there doesn't seem to be any other potential choices for difference between maps being able to represent $A$, and $B$, and they must thus be similair. Extremely vague ending of proof i know.
Is the "idea" of my proof correct, or does it need a more rigorous approach? Also, i do not want to prove it with jordan canonical form, because the block-subdiagonal canonical form is learned before jordan canonical form.