Connectedness of matrix conjugacy classes of a fixed real $A$ but with the first column of $A$ invariant 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.
 A: 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.
