This is a naive linear algebra question. I apologize for the level but I could not find an answer in the literature.
Let $A$ be a $n$ by $n$ matrix (say over $\mathbb C$). Suppose the Jordan form of $A$ is the matrix $J_\lambda$, for some partition $\lambda$ of $n$. My question is: how can one describe the set of invertible matrices which "jordanize" $A$? In other words, I would like to have an explicit description of $$G_A=\{M\in \textrm{GL}_n\,|\,M^{-1}AM=J_\lambda\}\subset \textrm{GL}_n.$$ I remember some time ago I thought the answer would be that $G_A$ is a product of $\mathbb C^\ast$, as many as there are distinct eigenvalues for $A$, which can be read off from $\lambda$. But now I was looking at a specific example more closely and I am unsure how one would go about proving that. Any help or reference is very much appreciated!
Added. For instance, if $n=2$, one can check by explicit computation that for the (already jordanized) matrix $$A= \begin{pmatrix} \eta & 0\\ 0 & \mu \end{pmatrix},\qquad \eta\mu\neq 0,\,\,\,\eta\neq \mu, $$ one has $$ G_A=\Big\{\begin{pmatrix} a & 0\\ 0 & b \end{pmatrix}:a,b\in\mathbb C^\ast\Big\}=\mathbb C^\ast\times \mathbb C^\ast. $$ I wonder if one can be as explicit as this (expressing $G_A$ in terms of $\mathbb C^\ast$ or maybe $\mathbb C$) for any matrix of given Jordan type.
Suppose that both $M$ and $N$ are in $G_A$. Then $$ (A =)\; MJM^{-1} = NJN^{-1} \implies\\ J = (M^{-1}N)^{-1}J(M^{-1}N) \implies\\ (M^{-1}N)J = J(M^{-1}N) $$ In other words: fix any one element $M \in G_A$. Let $$ C_J = \{X \in GL : XJ = JX\} $$ Then $G_A = \{MX: X \in C_J\}$. Note also that $C_J$ may be thought of as the kernel of the linear map $X \mapsto XJ - JX$.
Note: if $J$ is block diagonal with $$ J = \pmatrix{ J_1\\ &J_2\\ &&\ddots\\ &&& J_k } $$ Where each $J_i$ is a Jordan block, then a matrix will commute with $J$ if and only if it is block diagonal of the same shape and corresponding blocks commute.