Assume that $A \in M_{n}(\mathbb{R}) $ is a matrix. Then $A$ generates a one parameter (with parameter $t\in \mathbb{R}$) family of group automorphisms of $\mathbb{R}^{n}$ with $x\mapsto e^{tA}x,\;\;x\in \mathbb{R}^{n}$. Here $\mathbb{R}^{n}$ is counted as a group with the usual addition. So every matrix $A$ define an action of $\mathbb{R}$ on $\mathbb{R}^{n}$. with some abuse of notations, this action is denoted by $A$, again. The corresponding semi direct product gives us a Lie group $G_{A}=\mathbb{R}^{n} \rtimes_{A} \mathbb{R}$.
Is it true to say that $G_{A} \simeq G_{B}$, as two lie groups $\iff$ $A$ an $B$ are two similar matrices?
As abstract Lie groups, if you take $A=(2)$ and $B=(3)$ as $1 \times 1$ matrices, the Lie groups $G_A$ and $G_B$ are isomorphic, to the 2-dimensional nonabelian simply connected Lie group.