TITLE: When are two elements conjugated in GL(2), but not in SL(2)
QUESTION [13 upvotes]: Let $F$ be an arbitrary field. How can we describe the set of elements in $SL(2,F)$ which are conjugated in $GL(2,F)$ but not in $SL(2,F)$?
I would be happy already with a partial solution as given in the comments.

REPLY [8 votes]: There is a very general result on this kind of things, but I will have to dig it up when I get back to my office tomorrow. ADDED Here's the reference I promised. It's the beautiful article by G.E. Wall, Conjugacy classes in projective and special linear groups. Bull. Austral. Math. Soc. 22 (1980), no. 3, 339–364.
Anyway, in this particular case, I believe the following should be correct.
All of the conjugates of $A \in \operatorname{SL}(2,F)$ under $\operatorname{GL}(2,F)$ are also conjugate under $\operatorname{SL}(2,F)$ if and only if for each $G \in \operatorname{GL}(2,F)$ there is $S \in\operatorname{SL}(2,F)$ such that
$$
S^{-1} G^{-1} A G S = A
$$
that is, for each $G \in \operatorname{GL}(2,F)$ there is $S \in\operatorname{SL}(2,F)$ such that $$GS \in C_{\operatorname{GL}(2,F)}(A),$$ that is, in every coset of the centralizer $C_{\operatorname{GL}(2,F)}(A)$ there is an element of $\operatorname{SL}(2,F)$, that is $$\operatorname{GL}(2,F) = \operatorname{SL}(2,F) C_{\operatorname{GL}(2,F)}(A),$$ that is, the centralizer $C_{\operatorname{GL}(2,F)}(A)$ contains elements of arbitrary determinant.
In @BillCook's case, every centralizer contains the scalar matrices, which have arbitrary determinant, as all element are squares in $F$.
In @Ludolila's first case, the centralizer of $A$ is made of the elements of the form
$$
c I + d A = \begin{bmatrix}c&d\\-d&c\end{bmatrix}
$$
of determinant $c^2 + d^2 \ge 0$. So if you conjugate (as noted by Ludolila) $A$ by a matrix of negative determinant, you'll never be able to get back conjugating with an element in $\operatorname{SL}(2,F)$. In the second case, the centralizer contains all scalar matrices, so the determinant is arbitrary here.