When are two elements conjugated in GL(2), but not in SL(2) 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.
 A: 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.
A: Great question! Here are some thoughts:
First, if the field is closed under square roots (like $\mathbb{C}$), Bill already gave a great answer. 
What about $\mathbb{R}$?  For example, $ A= \left[
  \begin{array}{ c c }
     0 & 1 \\
     -1 & 0
  \end{array} \right]
$ and $(-A)$ are conjugated in $GL_2(\mathbb{R})$ but not in $SL_2(\mathbb{R})$. Indeed, a simple check shows that if $PAP^{-1}=-A$ then $\det P<0$. 
Another observation: even if for two matrices $A,B\in SL_2(\mathbb{R})$ we have $P\in GL_2(\mathbb{R})$ such that $PAP^{-1}=B$ and $\det P<0$, that still doesn't mean that they are not conjugated in $SL_2(\mathbb{R})$ by a different matrix. For example: $A= \left[
  \begin{array}{ c c }
     2 & 0 \\
     0 & \frac{1}{2}
  \end{array} \right]
$, $B= \left[
  \begin{array}{ c c }
     \frac{1}{2} & 0 \\
     0 & 2
  \end{array} \right]
$, and $P= \left[
  \begin{array}{ c c }
     0 & 3 \\
     4 & 0
  \end{array} \right]
$ . We can still find $P' \in SL_2(\mathbb{R})$ that conjugates $A$ and $B$.
So, what went wrong in the first example? Maybe it has something to do with the fact that $ A= \left[
  \begin{array}{ c c }
     0 & 1 \\
     -1 & 0
  \end{array} \right]
$ has a characteristic polynomial which is irreducible over $\mathbb{R}$?
