Restriction of scalars $G \mapsto \operatorname{Res}_{\mathbb C/\mathbb R} G$ is injective? Let $G$ be a linear algebraic group over $\mathbb C$, and let $G_0 = \operatorname{Res}(\mathbb C/\mathbb R, G)$ be the linear algebraic group over $\mathbb R$ obtained by Weil restriction of scalars.  One way to construct $G_0$ is by taking the defining polynomials for $G$ with complex coefficients, and rearranging these into a system of polynomials with real coefficients.  
Is the functor $G \mapsto G_0$ faithful?  It seems like this should be the case.  If we base change $G_0$ back to $\mathbb C$, we get the product $G \times G$, although not $G$ itself.  
In more algebraic terms, the question comes down to undoing this process of rearranging the system of polynomials.  If we have polynomials
$$f(T) = a_0 + a_1T + \cdots + a_nT^n \in \mathbb C[T]$$ 
then we write $T = X+iY$, $a_j = b_j + ic_j$, and rearrange the condition $f(t) = 0$ into the condition that two polynomials in the variables $X$ and $Y$ vanish.  The question is whether any two reversals of this process give the same complex algebraic variety over $\mathbb C$.  
 A: NB: I haven't checked all the details in this answer, so take it with a grain of salt.

I think what you're asking is the following:

If $\mathsf{Res}_{\mathbb{C}/\mathbb{R}}G_1\cong \mathsf{Res}_{\mathbb{C}/\mathbb{R}}G_2$ does that imply that $G_1\cong G_2$?

I think the answer is no if you just want to assume that $G_1$ and $G_2$ are linear algebraic groups. Namely, there exists unipotent groups $U$ over $\mathbb{C}$ which have no model over $\mathbb{R}$. Indeed, I think the construction given by [Sch, Proposition 2] produces a nilpotent Lie algebra over $\mathbb{C}$ which has no model (just take $d$ to be non-real) and then one can apply [Mil,Theorem 13.37(b)].
Consider then $U^\sigma:=U\times_{\mathrm{Spec}(\mathbb{C})}\mathbb{C}$ where the map $\text{Spec}(\mathbb{C})\to \text{Spec}(\mathbb{C})$ has global sections given by conjugation. Note that $U^\sigma\not\cong U$ as $\mathbb{C}$-groups else one could find a model of $U$ over $\mathbb{R}$ (e.g. by the theory in [Poo, §4.4]). That said, I claim that
$$\mathsf{Res}_{\mathbb{C}/\mathbb{R}}U\cong \mathsf{Res}_{\mathbb{C}/\mathbb{R}}U^\sigma$$
Indeed, it suffices to produce a $\Gamma_\mathbb{R}$-equivariant isomorphism
$$(\mathsf{Res}_{\mathbb{C}/\mathbb{R}} U)_\mathbb{C}\cong (\mathsf{Res}_{\mathbb{C}/\mathbb{R}} U^\sigma)_\mathbb{C}$$
But,
$$(\mathsf{Res}_{\mathbb{C}/\mathbb{R}} H)_\mathbb{C}\cong H\times H^\sigma$$
where the non-trivial element $\sigma$ in $\Gamma_{\mathbb{R}}$ is given by the map
$$\sigma:H\times H^\sigma\to H\times H^\sigma:(h,h')\mapsto (\sigma(h'),\sigma(h))$$
where we have
$$\sigma:H^\sigma\to H$$
by definition and
$$\sigma:H\to H^\sigma$$
is its inverse.
But, since $(U^\sigma)^{\sigma}\cong U$ it's not hard to see that
$$(\mathsf{Res}_{\mathbb{C}/\mathbb{R}} U)_\mathbb{C}\cong U\times U^\sigma\cong U^\sigma\times U\cong (\mathsf{Res}_{\mathbb{C}/\mathbb{R}} U^\sigma)_\mathbb{C}$$
in a $\Gamma_\mathbb{R}$-equivariant way.

If $G_i$ are reductive then I think it's true, but I haven't checked.
[Mil] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.
[Poo] Poonen, B., 2017. Rational points on varieties (Vol. 186). American Mathematical Soc..
[Sch] Scheuneman, J., 1967. Two-step nilpotent Lie algebras. Journal of Algebra, 7(2), pp.152-159.
