Splitting criterion for conjugacy classes in special linear group of prime degree over a finite field

Statement
Suppose $$n$$ is a prime number and $$q$$ is a prime power (the underlying prime of $$q$$ may or may not be equal to $$n$$). Then, there is a finite field $$\mathbb{F}_q$$ of size $$q$$, unique up to isomorphism. We denote the special linear group $$SL(n,\mathbb{F}_q)$$ by $$SL(n,q)$$. Similarly, we denote the general linear group $$GL(n,\mathbb{F}_q)$$ by $$GL(n,q)$$.

Our goal is to determine the splitting criterion for elements between $$GL(n,q)$$ and the normal subgroup $$SL(n,q)$$. More explicitly, we are interested in the question: given an element $$g \in SL(n,q)$$, under what conditions is the $$GL(n,q)$$-conjugacy class of $$g$$ the same as its $$SL(n,q)$$-conjugacy class? Further, if they are not equal, how many different $$SL(n,q)$$-conjugacy classes does the $$GL(n,q)$$-conjugacy class split into?

Case of no nth roots of unity
If $$\operatorname{gcd}(n,q - 1) = 1$$ (this is the same as saying that $$q$$ is not 1 mod $$n$$), then none of the conjugacy classes split. In fact, in this case, $$SL(n,q)$$ is a direct factor of $$GL(n,q)$$, hence a conjugacy-closed subgroup; for more, see isomorphism between linear groups when degree power map is bijective.

Case of primitive nth roots of unity
If $$\operatorname{gcd}(n,q - 1) = n$$ (this is the same as saying that $$q$$ is 1 mod $$n$$), then the only conjugacy classes that split are the ones that comprise a single Jordan block of size $$n$$:


 * There are $$n$$ such $$GL(n,q)$$-conjugacy classes in $$SL(n,q)$$. These correspond to eigenvalue choices that are the $$n$$ different $$n^{th}$$ roots of unity.
 * Each such $$GL(n,q)$$-conjugacy class splits into $$n$$ conjugacy classes in $$SL(n,q)$$.
 * Thus, there is a total of $$n^2$$ conjugacy classes in $$SL(n,q)$$ that we obtain after the splitting.