In dimension $d$ (anything $\geq 7$at least 7, there are continuous positive-dimensional (non-isotrivial) families of nilpotent Lie algebras in characteristic zero. I think)had previously heard this as folklore, and a positive dimensional moduli space but some searching yields Ming-Peng Gong's dissertation, which has explicit presentations. By the exponential correspondence, this yields families of unipotent groups of dimension $d$ defined over subfields of $\mathbb{Q}$. All real points in this moduli space yield real Lie groups \mathbb{R}$, all of whose real-analytifications have underlying manifolds that are diffeomorphic to$\mathbb{R}^d$. • A I will choose the family (147E), with generating basis$x_1,\ldots,x_7$, and nonzero brackets$[x_1,x_2] = x_4$,$[x_1,x_3]=-x_6$,$[x_1,x_5] = -x_7$,$[x_2,x_3]=x_5$,$[x_2,x_6]=\lambda x_7$,$[x_3,x_4]=(1-\lambda)x_7$, as$\lambda$varies over complex numbers satisfying$\lambda(\lambda-1) \neq 0$. Over the complex numbers, the Lie algebras are distinguished up to isomorphism by the value of$j(\lambda) = \frac{(1-\lambda+\lambda^2)^3}{\lambda^2(\lambda-1)^2}$. For each real$\lambda$, I will call the Lie algebra$L(\lambda)$, and the corresponding Lie group$G(\lambda)$. For any real quadratic extension$K/\mathbb{Q}$, and there is a Galois-conjugate pair$(\lambda, \lambda')$of distinct irrational elements, such that$K$-points in this moduli space j(\lambda) \neq j(\lambda')$. One reason is that are taken to each other $j^{-1}j$ is not equivariant under translation by one - you can find the nontrivial automorphism 6-element preimage of $K$.
• By fixing an embedding j(\lambda)$written out in Wikipedia's Modular lambda article. For such a pair,$K \L(\lambda)$is not isomorphic to \mathbb{R}$, we obtain two real algebraic groups over these two $K$-points that are non-isomorphic as real algebraic groups. The corresponding Lie groups are non-isomorphic L(\lambda')$, and$G(\lambda)$is not isomorphic to$G(\lambda')$as a real Lie groupsgroup. However, because the Lie algebras underlying groups of real points are not isomorphic. However, The isomorphism is given by transporting the nontrivial Galois automorphism of$K$induces through the functor$- \otimes_K \mathbb{R}$on Lie algebras, followed by exponentiation. This is highly discontinuous. For example, each$\exp(x_i) \in G(\lambda)$is taken to$\exp(x_i) \in G(\lambda')$, but$\exp (\lambda x_i) \in G(\lambda)$is taken to$\exp(\lambda' x_i) \in G(\lambda')$. In summary, we have two Lie groups$G(\lambda)$and$G(\lambda')$, we have an abstract group isomorphism on the underlying point-set$f$(by transport of Galois) between them, and a diffeomorphism$g$(because they are both diffeomorphic to$\mathbb{R}^7$) between them, but they are not isomorphic as Lie groups. 2 added 60 characters in body; added 11 characters in body I believe the following exist: 1. A positive integer$d$(anything$\geq 7$, I think), and a positive dimensional moduli space of unipotent groups of dimension$d$defined over$\mathbb{Q}$. All real points in this moduli space yield real Lie groups whose underlying manifolds are diffeomorphic to$\mathbb{R}^d$. 2. A real quadratic extension$K/\mathbb{Q}$, and a pair of distinct$K$-points in this moduli space that are taken to each other by the nontrivial automorphism of$K$. By fixing an embedding$K \to \mathbb{R}$, we obtain two real algebraic groups over these two$K$-points that are non-isomorphic as real algebraic groups, but are abstractly isomorphic. The corresponding Lie groups are non-isomorphic as real Lie groups, because the Lie algebras are not isomorphic. However, the nontrivial automorphism of$K$induces an isomorphism on the underlying point-set groups. 1 I believe the following exist: 1. A positive integer$d$(anything$\geq 7$, I think), and a positive dimensional moduli space of unipotent groups of dimension$d$defined over$\mathbb{Q}$. All real points in this moduli space yield real Lie groups whose underlying manifolds are diffeomorphic to$\mathbb{R}^d$. 2. A real quadratic extension$K/\mathbb{Q}$, and a pair of distinct$K$-points in this moduli space that are taken to each other by the nontrivial automorphism of$K$. By fixing an embedding$K \to \mathbb{R}$, we obtain two real algebraic groups over these two$K\$-points that are non-isomorphic as real algebraic groups, but are abstractly isomorphic. The corresponding Lie groups are non-isomorphic as real Lie groups, because the Lie algebras are not isomorphic.