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Let us work over finite fields $F_{p^k}$. Simulations seems to indicate:

Question 1: Consider a nilpotent matrix $A$, consider the set of all matrices $B$, such that $AB-qBA=0$, then cardinality of that set does not depend on $q\in F: q\ne 0$, is it true? So in particular for $A$ being a nilpotent, commute with $A$ and anticommute with $A$ — both have the same cardinality.

(Later edit: proved positively by Darij Grinberg).

Remarks: simulations show that the sets themselves depend on $q$ — only cardinalities are the same. Claim above is obviously not true without $A$ being nilpotent e.g. for $A=\operatorname{Id}$ everything commutes with $A$, but nothing non-zero anticommutes (except $\operatorname{char}(F)=2$).

Question 2: All the same as above seems to be true when $B$ restricted to be nilpotent, is it true? I.e. cardinalities of sets $$N_{q,A} = \{B: \text{$B$ nilpotent}, AB-qBA=0 \}$$ do not depend on $q\ne 0$ (for fixed $A$). Is it true?

Moreover if we stratify nilpotent cone by $B^k=0$, the same seems to be true for each strata: Question 3: I.e. $$N_{q,A,k} = \{ B: B^k=0 , AB-qBA=0 \}$$ the cardinality of these sets do not depend on $q\ne 0$ (for fixed $A$). Is it true?

Remark: in earlier version it was claimed that the sets themselves do not depend on $q$ for $k=2$ - but it is not true. ( As Darij Grinberg observed by argument and I rechecked simulation result - already for p=3, n=4 it is not true).

Link to Python simulations: https://www.kaggle.com/alexandervc/count-x-nilp-x-c-x-nilp-0

The case p=3, n=4 is considered in version 6 of the notebook, the table with results is just above the linked position: https://www.kaggle.com/alexandervc/count-x-nilp-x-c-x-nilp-0?scriptVersionId=75993799&cellId=10

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    $\begingroup$ The matrix $qA$ is similar to $A$ whenever $A$ is nilpotent and $q \neq 0$. (Indeed, this follows from the Jordan normal form, since each nilpotent Jordan block is similar to its own $q$-multiple.) Thus, $qA = CAC^{-1}$ for some invertible matrix $C$. Now, there is a bijection between the matrices $B$ satisfying $AB = qBA$ and the matrices $B$ satisfying $AB = BA$; namely, this bijection sends a matrix $B$ of the former type to $CB$. This answers Question 1. Not sure about Question 2. $\endgroup$
    – darij grinberg
    Commented Sep 30, 2021 at 0:38
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    $\begingroup$ Pretty sure Question 3 cannot have a positive answer. If the sets are equal, then in particular any matrix $B$ satisfying $AB = BA$ and $B^2 = 0$ must satisfy $AB = 2BA$ and therefore $AB = 0$. But it is easy to find a situation where this is violated (e.g., a quotient algebra $\mathbb{F}_3\left[a,b\right]/\left(a^3, b^2\right)$ in its regular representation as matrices). Of course, this doesn't preclude the possibility that the cardinalities might again be equal. $\endgroup$
    – darij grinberg
    Commented Sep 30, 2021 at 0:46
  • $\begingroup$ The equality of sizes in Question 3 seems to hold even for $q = 0$. $\endgroup$
    – darij grinberg
    Commented Sep 30, 2021 at 1:09
  • $\begingroup$ Actually, the equality of sizes in Question 1 seems to hold for $q = 0$ as well, though this would be far from obvious. $\endgroup$
    – darij grinberg
    Commented Sep 30, 2021 at 1:14
  • $\begingroup$ Ah no, the equality of sizes in Question 1 fails for $q = 0$ when $A = \begin{pmatrix} 0&1&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}$ and $F = \mathbb{F}_2$. And the same counterexample works for Question 3. So $q = 0$ is not an option. $\endgroup$
    – darij grinberg
    Commented Sep 30, 2021 at 1:19

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