Consider

All you need is love:-)) Not really, just take a semidirect product family of complex nilpotent Lie algebras $G_{a,b}={\mathbb R}\ltimes {\mathbb R}^2$L_a$ depending on a complex parameter $a$. Such a family exists in dimension 7, or maybe even 6 (I do not remember the nilpotent classification).The action is diagonal with two eigenvalues

Now consider a wild automorphism of complex numbers $a$ \phi$. Then $L_a$ and $b$. Its L_{\phi (a)}$ for a generic $a$ are isomorphic as Lie group isomorphism class is determined by algebras over $a/b$.

Not let {\mathbb Q}$ but not over $\phi : {\mathbb R}\rightarrow {\mathbb {\mathbb R}$ be an isomorphism of or ${\mathbb Q}$-vector spacesC}$. Then $G_{a,b}$ Love is obviously all you need((-: the corresponding simply connected Lie groups are isomorphic to $G_{\phi(a),\phi(b)}$ as abstract groups but $\phi(a)/\phi(b)$ could be easily different from $a/b$.not as Lie groups.

Consider a semidirect product of $G_{a,b}={\mathbb R}\ltimes {\mathbb R}^2$. The action is diagonal with two eigenvalues $a$ and $b$. Its Lie group isomorphism class is determined by $a/b$.

Not let $\phi : {\mathbb R}\rightarrow {\mathbb R}$ be an isomorphism of ${\mathbb Q}$-vector spaces. Then $G_{a,b}$ is obviously isomorphic to $G_{\phi(a),\phi(b)}$ as abstract groups but $\phi(a)/\phi(b)$ could be easily different from $a/b$.