This question is related to the question I asked but the underlying field is $\mathbb C$ instead of $\mathbb R$.
Let $A \in M_n(\mathbb R)$ be a fixed real matrix. The set $\{S^{-1} A S: S \in GL_n(\mathbb R)\}$ is a continuous image of $GL_n(\mathbb R)$ defined by $\phi: GL_n(\mathbb R) \ni S \mapsto S^{-1}AS$, so I think it will have at most two connected components corresponding to general linear maps with positive determinants and negative determinants. Let $A = (a_1, \dots, a_n)$ where $a_j \in \mathbb R^n$ denotes the columns of $A$. Let $$F = \{S^{-1} A S: S \in GL_n(\mathbb R) \text{ and } (S^{-1}AS)_{\cdot,1} = a_1\}.$$ The condition $(S^{-1} A S )_{\cdot,1} = a_1$ is equivalent to $(AS-SA) e_1 = 0$ where $e_1 = (1, 0, \dots, )$. This is also a linear equation over $S$. So if we define $\psi: M_n(\mathbb R) \to \mathbb R^n$ by $S \mapsto (AS-SA)e_1$. Then Equivalently, $$F = \{R^{-1}AR: R \in \text{ker}(\psi) \cap GL_n(\mathbb R)\},$$ where $\text{ker}(\psi)$ is a linear subspace in $M_n(\mathbb R)$.
My question is: How many connected components does the set $F$ have? Is it at most two connected components?
Edit 1: Let $E_{\pm} = \{S \in GL_n(\mathbb R): (AS-SA)e_1 = 0, \pm \det(S) > 0\}$. Let $\phi: GL_n(\mathbb R) \to M_n(\mathbb R)$ be defined by $S \mapsto S^{-1}AS$. As discussed in the comments, if $n$ is odd, then $\phi(E_+) = \phi(E_{\_})$ which says the image does not differentiate the sign of determinants. As demonstrated by amsmath, $E_+$ can have more than $1$ connected component, but is it possible that all the components are mapped to the same connected component by $\phi$?
Edit 2: This question Connectedness of matrix conjugacy class might be helpful here. It concerns whether $\{S^{-1}AS: S \in GL_n(\mathbb R)\}$ is connected (we know it has at most two connected components).
p.s. The answer below provides nice insights but not addressing this particular question.
The answer to your question is no, not necessarily. To see this, let $A$ be such that $\ker A = \operatorname{span}\{e_1\}$. Then the condition $(SA-AS)e_1 = 0$ reduces to $Se_1\in\operatorname{span}\{e_1\}$, that is, $Se_1 = \lambda e_1$ for some $\lambda\in\mathbb R$. The sets $$ E_\pm := \{S\in GL(n,\mathbb R) : Se_1\in\operatorname{span}\{e_1\},\;\pm\det S > 0\} $$ are disconnected. So, the number of connected components of $E := E_+\cup E_-$ is at least two. We will now show that $E_+$ is not connected. For this, choose some $S\in GL(n,\mathbb R)$ with $Se_1 = -e_1$ and $\det S > 0$ and assume that $t\mapsto S(t)$, $z\in [0,1]$, is a continuous path from $S = S(0)$ to the identity $I = S(1)$ in $E_+$. Then for each $t\in [0,1]$ we have $S(t)e_1 = \lambda(t)e_1$ for some $\lambda(t)\in\mathbb R$, where $\lambda(0) = -1$ and $\lambda(1) = 1$. But as $\lambda(t) = \langle S(t)e_1,e_1\rangle$ is continuous, by the intermediate value theorem there exists some $t_0\in (0,1)$ such that $\lambda(t_0) = 0$, i.e., $S(t_0)e_1 = 0$, which is a contradiction.