This is an eccentric question: recall that a smooth Lie group structure on $\mathbb R^n$ is uniquely identified by a triple $(\mu,\iota,e)$ where $\mu:\mathbb R^n\times\mathbb R^n\to\mathbb R^n$ is the multiplication, $e\in\mathbb R^n$ is the neutral element, and $\iota:\mathbb R^n\to\mathbb R^n$ is the inversion. Since we can canonically identify the tangent space $T_e\mathbb R^n$ with $\mathbb R^n$, for every triple $(\mu,\iota,e)$ we obtain (by differentiation) a Lie bracket as a bilinear map $\beta:\mathbb R^n\times\mathbb R^n\to\mathbb R^n$. My question is the following: do there exist two triples $(\mu,\iota,e)$ and $(\mu',\iota',e)$ (note: same neutral element) such that $\mu\not=\mu'$ but $\beta=\beta'$?

I conjecture that using some nilpotent Lie groups one can construct an example but that's only mere speculation.