Let us define $\phi : GL(n,\Bbb R)\to C_A$, $\;\phi(S) = SAS^{-1}$ and the sets $$ E_\pm := \{S\in GL(n,\Bbb R) : \pm\det S > 0\}. $$ If $n$ is odd, then $\phi(E_+) = \phi(E_-) = C_A$ (because $\phi(-S) = \phi(S)$) and hence $C_A$ is connected. But what about even $n$?
EDIT: I just saw that
Lemma 1. If $\det A < 0$, then still $\phi(E_+) = \phi(E_-)$ and hence $C_A$ is connected.
Proof. Indeed, if $T\in\phi(E_-)$, $T = \phi(S_0)$, $\det S_0 < 0$, then $\det(S_0A) > 0$ and $T = \phi(S_0A)\in\phi(E_+)$. The other inclusion is proved similarly.
So the question reduces to $n$ even and $\det A\ge 0$.
EDIT2: Here is another fact.
Lemma 2. If there is $S_0\in E_-$ that commutes with $A$, then $C_A$ is connected.
Proof. Let $T\in C_A$. Let us show that we can find a path within $C_A$ from $T$ to $A$. Let $T = SAS^{-1}$ with $S\in GL(n,\Bbb R)$. If $S\in E_+$, we find a path from $S$ to $I$ in $E_+$ and hence a path in $C_A$ from $T$ to $A$. If $S\in E_-$, we find a path within $E_-$ from $S$ to $S_0$. Its image under $\phi$ is again a path from $T$ to $A$ within $C_A$.
EDIT 3: For arbitrary $\lambda\in\Bbb R$ we have $C_{A-\lambda I} = C_A - \lambda I$. As this is just a translation in $\Bbb R^{n\times n}$ of $C_A$ by $\lambda I$, it follows that $C_A$ is connected if and only if $C_{A-\lambda I}$ is connected.
Therefore we can conclude the following: Let $J$ be the real Jordan form of $A$. Then $C_A = C_J$. If $A$ has a real eigenvalue $\lambda_0$ which appears in $J$ in a $k\times k$ Jordan block with $k$ odd, then $C_A$ is connected. Indeed, due to the above, we can shift $A$ and $J$ simultaneously and thus assume that $A$ and $J$ are invertible. Let $\tilde J$ be $J$, but with $-\lambda_0$'s instead of $\lambda_0$'s on the diagonal of the $k\times k$ Jordan block. Then $\tilde J$ commutes with $J$ and hence so does $\tilde JJ$. Since $\det(\tilde JJ) < 0$, $C_J = C_A$ is connected by Lemma 2.
We summarize for the critical matrices: In the real Jordan form each Jordan block corresponding to a real eigenvalue has size $k\times k$ with $k$ even.
I conjecture that the following are equivalent:
- $C_A$ is connected
- There exists $S\in E_-$ that commutes with $A$.
- There exists a Jordan block $J$ of $A$ for which $C_J$ is connected.
- There exists a real odd-sized Jordan block of $A$.
I could only prove (4)$\Rightarrow$(3), (3)$\Rightarrow$(1), and (2)$\Rightarrow$(1) so far.
Remark: This question is related to and motivated by Connectedness of matrix conjugacy classes of a fixed real $A$ but with the first column of $A$ invariant
The question boils down to: do there exist (for even $n>0$) matrices in $E_+$ whose centraliser in $GL(n,\Bbb R)$ is entirely contained in $E_+$ (in other words your (1) and (2) are equivalent). To see the direction you did not address in the question, remark that $\phi(E_+)$ is certainly a connected component of$~C_A$; if $C_A$ is connected, then it is all of$~C_A$. But then for any chosen $S\in E_-$ one has $\phi(S)\in\phi(E_+)$, say $\phi(S)=\phi(T)$ with $T\in E_+$ whence $T^{-1}S\in E_-$ centralises $A$.
Indeed such matrices exists; any maximal size Jordan block will do. The only matrices that commute with them are polynomials in them, and when invertible such matrices have positive determinant.
As for your other characterisations of the situation, it would seem to me that a disconnected conjugacy class can happen ever in the presence of imaginary eigenvalues (try a block triangular $4\times4$ matrix with equal rotation matrices as its $2\times2$ diagonal block).