Distinguish smooth affine algebraic group by its underlying variety If $G_1, G_2$ are two connected affine smooth algebraic groups over an algebraically closed field $k$, and $G_1 \cong G_2$ as algebraic varieties, must they be isomorphic as algebraic groups? 
Although connected unipotent groups of same dimension all have same underlying varieties (i.e the affine space), I suspect that with suitable reductive assumptions (e.g one of them is reductive + other conditions), $G_1$ and $G_2$ must be isomorphic as algebraic groups.
There is a classification of connected reductive groups, but do the topological invariants (e.g fundamental group + dimension + cohomology) determine the group?
What if $k$ is only assumed to be a perfect field?
Example: the dimension $1$ case and semisimple rank $1$ case are easy by classification. For instance, $SL_2$ and $PGL_2$ are different because they have different fundamental groups.
Edit: another way of saying the question is: how to characterize those $G$ on which there exists only one / finitely many smooth (reductive) algebraic group structures? If one is reductive, must the other be?
 A: Let's consider the last version of the question over $\mathbb C$. We will prove two facts: I, II.
I. If two complex algebraic groups $G_1$ and $G_2$ are diffeomorphic, then they are either both reductive or both non reductive.
II. For any reductive complex algebraic group there exist at most finite number of diffeomorphic to it complex algebraic reductive groups. 
Obvious remark. Of course, two complex Lie groups that the same as algebraic varieties are diffeomorphic.  
Proof of I. This follows from the following claim.
Claim. A complex algebraic $n$-dimensional group  is reductive if and only if it is homotopy equivalent to an (orientable) real compact manifold of dimension $n$.
Clearly, this shows that a non-reductive group can not be even diffeomorphic to a reductive one.
Let's see how to prove this claim. We will use two alternative characterisations of reductive groups.
First, a complex algebraic group is reductive if and only if it has a compact real form, see for example here:
http://www.math.uchicago.edu/~mbergeron/ComplexReductive.pdf
If follows from this, that any complex reductive group is homotopy equivalent to a compact manifold of half dimension, its real compact form. 
See, for example, "Other characterisations" here: https://en.wikipedia.org/wiki/Reductive_group .
For example $(\mathbb C^*)^n$ is homotopy equivalent to $(S^1)^n$. In particular $H_n(G,\mathbb Z)$ of a complex reductive group $G$ of dimension $n$ is $\mathbb Z$. 
Second. The mid-dimensional homology group vanishes for non-reductive groups, even more, non-reductive groups are homotopy equivalent to compact manifolds of dimensions less than the half. To prove this one uses the following definition of reductive groups: A non-reductive group has a nontrivial normal unipotent subgroup. 
See here https://en.wikipedia.org/wiki/Reductive_group.
Now, if we quotient a group by a normal unipotent subgroup, the homotopy type doesn't change. So we can quotient until we get a reductive group. 
Proof of II. If is easy to see from the proof of I, that if $G_1$ is diffeomorphic to $G_2$, then their (half-dimensional) compact forms are homotopy equivalent. So we just need to know that in each dimension there exists only finite number of connected compact Lie groups. This is indeed well known, see for example the references in the following mathoverflow answer: Classification of (compact) Lie groups
A: $G_1:=GL(2)(\cong SL(2)\rtimes \mathbb G_m)$ is isomorphic as a variety to $G_2:=SL(2) \times \mathbb G_m$ via the map
$$
A\mapsto \Big(\big(\begin{smallmatrix}\det(A)^{-1} & 0 \\ 0 & 1 \end{smallmatrix}\big)A , \det(A)\Big).
$$
We will show that $G_1$ and $G_2$ (which are both reductive) are not isomorphic as algebraic groups over fields of characteristic 0. 
This is not immediate, as for $k$ a perfect field of characteristic 2 there is an isomorphism of abstract groups $\psi: GL(2)(k) \to (SL(2) \times \mathbb G_m)(k)$ given by $\psi(A) = (\det(A)^{-\frac{1}2}A, \sqrt{\det(A)})$. 
So suppose $\varphi: G_1 \to G_2$ is an isomorphism. 
First proof: Note that $Z(G_1)=\mathbb G_m$ and $Z(G_2)=C_2 \times \mathbb G_m$. Then $\varphi$ induces an isomorphism $Z(G_1) \cong Z(G_2)$, which is absurd since over a field with characteristic different from 2 the equation $x^2=1$ has 2 solutions in $\mathbb G_m$ and 4 solutions in $C_2 \times \mathbb G_m$. 
Second proof: 
As $\mathbb G_m$ is diagonalizable and $\varphi^{-1}$ preserves this property, so is $\varphi^{-1}(\mathbb G_m)$ and hence is contained in a maximal torus, which are all conjugate. So $\psi:= \varphi^{-1}|_{\mathbb G_m}$ gives rise to a so-called cocharacter $\psi \in Hom(\mathbb G_m, T)$ for $T$ the maximal torus of diagonal matrices of $GL_2$. By standard facts about cocharacters, $\psi$ has the form $z\to (z^a, z^b)$ for some integers $a, b$. Since the domain of $\psi$ is central in $G_2$, so must be $\text{im } \psi$, which forces $a=b$. Since $\psi$ is injective on $\bar{k}$-rational points, we have $a=\pm 1$. Then $\psi(-1)=(-1, -1)$.
On the other hand, $\varphi^{-1}$ affords a 2-dimensional algebraic representation of $SL_2$, so $\varphi^{-1}(SL_2)=SL_2$. This finally gives a contradiction as the images of $\varphi^{-1} (SL_2)$ and $\varphi^{-1}(\mathbb G_m)$ are not disjoint. 
