Can we classify the general linear maps such that for a fixed matrix $A \in M_n(\mathbb R)$ the conjuagation action fixes the first column? 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 $\mathcal E = \ker(\phi) \cap GL_n(\mathbb R)$.
 A: 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
