We regard an isomorphism of Lie groups to mean a group isomorphism which is simultaneously a diffeomorphism of the underlying smooth manifold. I'm wondering about how much rigidity is imposed by this definition.
Question: If we have maps $f, g: G\rightarrow H$, where $G, H$ are Lie groups, $f$ is an abstract group isomorphism, and $g$ is a diffeomorphism, must $G$ and $H$ be isomorphic as Lie groups?
I think there should be a (possibly easy) counterexample, but neither I nor the professors I've asked could immediately find one.
EDIT: assume $G$ and $H$ are connected.
Best Answer
The answer to your question is "no". I've rewritten my previous answer to include details.
In dimension $d$ at least 7, there are continuous positive-dimensional (non-isotrivial) families of nilpotent Lie algebras in characteristic zero. I had previously heard this as folklore, but some searching yields Ming-Peng Gong's dissertation, which has explicit presentations. By the exponential correspondence, this yields families of unipotent groups over subfields of $\mathbb{R}$, all of whose real-analytifications have underlying manifolds that are diffeomorphic to $\mathbb{R}^d$.
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 $K/\mathbb{Q}$, there is a Galois-conjugate pair $(\lambda, \lambda')$ of irrational elements, such that $j(\lambda) \neq j(\lambda')$. One reason is that $j^{-1}j$ is not equivariant under translation by one - you can find the 6-element preimage of $j(\lambda)$ written out in Wikipedia's Modular lambda article. For such a pair, $L(\lambda)$ is not isomorphic to $L(\lambda')$, and $G(\lambda)$ is not isomorphic to $G(\lambda')$ as a real Lie group.
However, the underlying groups of real points are isomorphic. The isomorphism is given by transporting the nontrivial Galois automorphism of $K$ 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 $f$ (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.