This question out of curiosity is inspired by a Hoffman & Kunze exercise:
(Section 7.3, Excercise 11) Let $N_1$ and $N_2$ be $6\times 6$ nilpotent matrices over the field F that have the same minimal polynomial and the same nullity. Prove that they are similar and show that this is not true for $7\times 7$ nilpotent matrices.
Linear algebra tells us that this question is equivalent to why 6 and 7 has and doesn't have, respectively, the property described in the title. (degrees of minimal polynomial and nullity correspond to the arm and leg length of the partition of the matrix's sole eigenvalues' algebraic multiplicity).
In this case, we can simply enumerate all partitions and get the result (in particular, 3+3+1=3+2+2), but I'm interested in the criterion for detecting numbers that behave just like 6. I have tried using the divisibility of succeeding squared number (for 6 and 7, 3, where 3|6 and 3$\nmid$7) but this path doesn't seem to be fruitful (15 do have this property but 4$\nmid$15) (sad face). I feel like this has a simple solution but I just can't think of any ways to arrive at it.
By the way, this is my first post on the network!!! Please let me know how I can improve my questions, no matter whether it is about the length/depth/politeness/clearness.