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Let $A \in GL_n(\mathbb R)$ be fixed. Let us consider the conjugation action by $G \in GL_n(\mathbb R)$, i.e., $G^{-1}AG$. I would like to see a way to identify the matrices such that the action fixes the first column of $A$. That is, what can we say about the set \begin{align*} \mathcal E = \{G \in GL_n(\mathbb R): G^{-1}AGe_1 = a_1\}, \end{align*} where $e_1$ is the standard basis vector and $a_1$ is the first column of $A$. In particular, is the set connected? Let us exclude the trivial case: $A$ is a multiple of $I$ or $-I$.

If we define a linear map $\phi: M_n(\mathbb R) \to \mathbb R^n$ by $X \mapsto (XA-AX)e_1$, then $\mathcal E = \ker(\phi) \cap GL_n(\mathbb R)$.

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One can not say that $\mathcal E$ is always connected.

For $n=3, $(and similarly $n>3$) let $A$ be a matrix with $a_{i1}=1,\quad \forall i \in \{1,2,\ldots,n\} $

Then $\mathcal E$ contains the identity matrix whose determinant is $1$ and it also contains the following matrix with determinant $-1$:

$$\begin{pmatrix} 1&0&0\\0&0&1\\0&1&0 \end{pmatrix}$$

According to the previous version of your question about the (semi)group structure of $\mathcal E$, observe that if you can prove that $\mathcal E$ is a semigroup then it is a group, too. Because the inverse of a every matrix $G$, can be written in the form of a polynomial in $G$.

I think that $\mathcal E$ is not a Lie group because it is contained in a sub vector space $F$, the kernel you mentioned, whose dimension is equal to the dimension of $G$. Moreover $Exp(F)$ is contained in $\mathcal E$ But in general case, $F$ is not a Lie algebra with the commutator Lie bracket. Please see the comment by Corbennick

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    $\begingroup$ I was in mind probably $\mathcal E$ would have two (or more) components corresponding to the positive determinant and negative determinant. Would you comment on how many components would $\psi(\mathcal E)$ have, where $\psi: B \mapsto B^{-1}AB$ provided $A$ has a real eigenvalue? In this case, the positive and negative determinant would be mapped into the same image. $\endgroup$
    – user1101010
    Commented Oct 18, 2018 at 16:29
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    $\begingroup$ @user9527 In fact one can generalize your interesting question in the following form: is there a linear subvector space of $M_n(\mathbb{R}$ which intersect $GL_n(\mathbb{R}$ in more than 2 components?Is there an example of this situation with codimension is equal $n$?(In your question that kernel has codimension n) $\endgroup$
    – Ali Taghavi
    Commented Oct 18, 2018 at 16:54
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    $\begingroup$ In $2\times 2$ matrices the matrices of the form $[[x,y],[y,x]]$ intersect $GL_2$ in the subset of the plane with equation $xy\neq 0$, which has 4 components. $\endgroup$
    – YCor
    Commented Oct 18, 2018 at 17:04
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    $\begingroup$ @user9527 Yes. In size $n$, in the subspace of upper triangular matrices, there are $2^n$ components of invertible matrices. $\endgroup$
    – YCor
    Commented Oct 18, 2018 at 17:06
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    $\begingroup$ So, $G\to F$ is an immersion between manifolds of the same dimension, so has an open image; $F$ is the the lie algebra of $G$. $\endgroup$
    – YCor
    Commented Oct 19, 2018 at 12:19

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