Are the groups $SU(2, ℂ)$ and $U(1, ℍ)$ isomorphic? Both these groups are double covers of $SO(3, ℝ)$ so they must share the same Lie Algebra at least. 
More generally if 2 Lie groups share a Lie algebra, and both are N-covers of another group, are the two groups isomorphic? 
 A: $SU(2)$ and $U(1, \mathbb{H})$ are isomorphic. However, it is not true that two Lie groups having isomorphic Lie algebras and being $N$-covers of another Lie group are isomorphic. For example, both $SU(2)\times SO(3)$ and $SO(4)$ are double covers of $SO(3)\times SO(3)$, and have Lie algebras isomorphic to $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, but they are not isomorphic. 
A: To have an explicit isomorphism, consider $\Bbb H$ with the usual basis $1,i,j,k$ over $\Bbb R$, instead as a vector space over $\Bbb C$ with basis $1,j$. Then the quaternion $q:=(x+iy)+j(z+iw)$, $x,y,z,w\in\Bbb R$, acts by  left multiplication  on the basis elements $1$, respectively $j$, as the row components of the matrix
$$
A(q)
=
\begin{bmatrix}
x+iy & z+iw\\
-z+iw & x-iy
\end{bmatrix}
$$
show it. So it induces a $\Bbb C$-algebra homomorphism from $\Bbb H$ to $M_2(\Bbb C)$, the algebra of $2\times 2$ matrices with complex components. The squared norm of a quaternion corresponds to the determinant, 
$$
|q|^2 =x^2+y^2+z^2+w^2 = (x+iy)(x-iy)-(z+iw)(-z+iw)=\det A(q)\ ,
$$
so the norm one quaternions (versors) correspond to elements in $\operatorname{SU}(2,\Bbb C)$. The (restriction of the) map $A$ defines the needed isomorphism.
A simplest counterexample would be the $2:1$ cover of $S^1=\{\ z\in\Bbb C\ :\ |z|=1\ \}$ given elementwise by $z\to z^2$, and the disconnected cousin $\{\pm 1\}\times S^1$.
