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 $\mathcal E = \ker(\phi) \cap GL\_n(\mathbb R)$.



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?
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](https://mathoverflow.net/questions/313136/a-lie-group-which-is-sat-in-its-lie-algebra)


