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.

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 $$N_{q,A} = \{ \text{$B$ nilpotent}, AB-qBA=0 \}$$ do not depend on $q\ne 0$ (for fixed $A$). Is it true?

Question 3: The stronger might be true if impose $B^2=0$: not only cardinality, but the sets themselves do not depend on $q$ (for fixed $A$). I.e. $$N2_{q,A} = \{ B^2=0 , AB-qBA=0 \}$$ these sets do not depend on $q\ne 0$ (for fixed $A$). Is it true?

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

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

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.

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=Id everything commute with A, but nothing non-zero anticommute (except $char(F)=2$).

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 $$N_{q,A} = \{ B-nilpotent , AB-qBA=0 \}$$ do not depend on $q\ne 0$ (for fixed $A$). Is it true  ?

Question 3: The stronger might be true if impose $B^2=0$: not only cardinality, but the sets themselves do not depend on $q$ (for fixed $A$) I.e. $$N2_{q,A} = \{ B^2=0 , AB-qBA=0 \}$$ these sets do not depend on $q\ne 0$ (for fixed $A$). Is it true  ?

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

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.

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 $$N_{q,A} = \{ \text{$B$ nilpotent}, AB-qBA=0 \}$$ do not depend on $q\ne 0$ (for fixed $A$). Is it true?

Question 3: The stronger might be true if impose $B^2=0$: not only cardinality, but the sets themselves do not depend on $q$ (for fixed $A$). I.e. $$N2_{q,A} = \{ B^2=0 , AB-qBA=0 \}$$ these sets do not depend on $q\ne 0$ (for fixed $A$). Is it true?

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

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 $q\ne 0$, is it true ? So in particular for $A$ being a nilpotent , commute with $A$ and anticommute with $A$ - both have the same cardinality.

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=Id everything commute with A, but nothing non-zero anticommute (except $char(F)=2$).

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 $$N_{q,A} = \{ B-nilpotent , AB-qBA=0 \}$$ do not depend on $q\ne 0$ (for fixed $A$). Is it true ?

Question 3: The stronger might be true if impose $B^2=0$: not only cardinality, but the sets themselves do not depend on $q$ (for fixed $A$) I.e. $$N2_{q,A} = \{ B^2=0 , AB-qBA=0 \}$$ these sets do not depend on $q\ne 0$ (for fixed $A$). Is it true ?

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

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.

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=Id everything commute with A, but nothing non-zero anticommute (except $char(F)=2$).

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 $$N_{q,A} = \{ B-nilpotent , AB-qBA=0 \}$$ do not depend on $q\ne 0$ (for fixed $A$). Is it true ?

Question 3: The stronger might be true if impose $B^2=0$: not only cardinality, but the sets themselves do not depend on $q$ (for fixed $A$) I.e. $$N2_{q,A} = \{ B^2=0 , AB-qBA=0 \}$$ these sets do not depend on $q\ne 0$ (for fixed $A$). Is it true ?

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