Counterexample to "kernel determines image" Working over a base field, there is a typical homomorphism theorem for affine algebraic groups ensuring that any two homomporphisms $G \to H_1$, $G \to H_2$ with the same kernel in $G$ have isomorphic images. Proving this makes heavily use of the nontrivial fact that injective homomorphisms between Hopf algebras over fields are faithfully flat.
Working over general base rings (which are not fields), it is easy to find non-faithfully flat injective Hopf algebra homomorphisms: For example over the base ring of integers, $\mathbb{Z}[X] \to \mathbb{Z}[X]$ sending $X$ to $nX$ for any $n \not\in \{0,1,-1\}$ is not faithfully flat. (It corresponds to the multiplication-by-$n$-homomorphism $\mathbb{G}_a \to \mathbb{G}_a$.) I suspect the above statement fails in this general setting, but I cound not find any counterexample so far.
Question 1: Are there homomorphisms $f_1 \colon G \to H_1$, $f_2 \colon G \to H_2$ of affine algebraic groups with the same kernel such that $\mathrm{Im}(f_1)$ is not isomorphic to $\mathrm{Im}(f_2)$?
More specifically:
Question 2: Is there an injective homomorphism $f \colon G \to H$ of affine algebraic groups such that $\mathrm{Im}(f)$ is not isomorphic to $G$?

Edit 1
Here is an attempt to find a counterexample to question 2. My idea is to adapt known homomorphisms of algebraic groups which can be defined over general base rings but show different behavior after base extensions to fields of different characteristics.
For any base ring $R$ there is the circle group $C$ defined for any $R$-algebra $A$ by
$$ C(A) = \{ (x,y) \in A^2 : x^2+y^2 = 1 \}. $$
The identity element is $(1,0)$, and the group operations are given by
$$ (x_1,y_1) \cdot (x_2,y_2) = (x_1x_2 - y_1y_2, x_1y_2+ x_2y_1), \text{ and } (x,y)^{-1} = (x,-y). $$
If $R$ has an element $i \in R$ with $i^2 = -1$, there is a homomorphism
$$ \varphi \colon C \to \mathbb{G}_m, \quad (x,y) \mapsto x+iy.$$
This morphism is quite exiting: If $2$ is invertible in $R$ then $\varphi$ is already an isomorphism. If $2$ not not invertible however, then $\varphi$ is not injective, and we can extend $R$ to a field of characteristics 2, where $\varphi$ is neither injective nor surjective (it maps onto $\mu_2$ then). This example can be found in [Waterhouse, Introduction to affine group schemes, chapter 1, exercise 11].
Since $\varphi$ has nontrivial kernel, it cannot be a counterexample to (2). So let's add an additional equation:
Let $C'$ be the subgroup of the circle group given by
$$ C'(A) = \{ (x,y) \in A^2 : x^2+y^2=1, 3y=0 \}. $$
Now the restriction of $\varphi$ to $C'$ has trivial kernel, but it is not surjective anymore - we have to add at least one additional equation $3t^2=3$ to $\mathbb{G}_m$. That is, $\varphi$ induces a homomorphism of algebraic subgroups
$$ \{ (x,y) : x^2+y^2=1, 3y = 0 \} \to \{ t : t \text{ invertible}, 3t^2=3 \}. $$
I think this might be a counterexample to (2) since it becomes an isomorphism after extending the base ring to any field, but I cannot see how it should be an isomorphism over (say) the Gaussian integers $R = \mathbb{Z}[i]$. However, I also cannot see at the moment why it should be (scheme-theoretically) surjective.
To restate the question in terms of Hopf algebras: There is a homomorphism
$$ (\mathbb{Z}[i])[T,T^{-1}]/(3(T^2-1)) \to (\mathbb{Z}[i])[X,Y]/(X^2+Y^2-1, 3Y), \quad T \mapsto X+iY. $$
Is this morphism injective but not surjective? (Of course (2) asks for a little bit more: Is the left hand side not ismomorphic to the right hand side?)

Edit 2
Unfortunately, this specific morphism is already an isomorphism which can be easily seen by condiering the localizations of $\mathbb{Z}[i]$ at all primes $\mathfrak{p}$. For all primes with $3 \notin \mathfrak{p}$ it is quite obvious that the morphism is basically the same as the identity map $\mu_2 \to \mu_2$, since $3$ is invertible in $\mathbb{Z}[i]_\mathfrak{p}$. On the other hand, if $3 \in \mathfrak{p}$ then $2$ is invertible in $\mathbb{Z}[i]_\mathfrak{p}$. In this case, $\varphi \colon C \to \mathbb{G}_m$ is an isomorphism, and it can be easily checked that its inverse maps $\{ t : t \text{ invertible }, 3t^2=3 \}$ to $C'$.
In order to get a counterexample in the way sketched above, we have to choose some other equation (instead of $3y=0$) which nontrivially affects the groups over the base ring $\mathbb{Z}[i]_\mathfrak{p}$ for $\mathfrak{p} = (1+i)$.
 A: Basically, I asked about the existence of a non-closed affine subgroup scheme not isomorphic to its closure (in some affine group scheme). I started by investigating a morphism between the circle group and the multiplicative group which behaves strange under base change. Finally, I found a desired example in that way:
Let's begin with the same situation as above: Let $R = \mathbb{Z}[i]$ be the ring of Gaussian integers, and let $C = \{ (x,y) : x^2+y^2=1 \}$ be the circle group over $R$. We have a homomorphism
$$ \varphi \colon C \to \mathbb{G}_m, \quad (x,y) \mapsto x+iy $$
which becomes an isomorphism after base change to any ring containing $1/2$. Over such rings, the inverse morphism is given by
$$ \mathbb{G}_m \to C, \quad t \mapsto \left( \frac{t+t^{-1}}{2}, \frac{t-t^{-1}}{2i} \right). $$
As mentioned above, $\varphi$ has a nontrivial kernel over the base ring $R$, so $C$ cannot be considered a subgroup of $\mathbb{G}_m$.
Let's consider the subgroup $C_3 = \{ w \in C : w^3 = 1 \}$ of all elements of order dividing 3 instead. It is clear that $\varphi$ restricts to a homomorphism
$$ \varphi_3 \colon C_3 \to \mu_3, $$
where $\mu_3 = \{ t : t^3 = 1 \}$ is the group of third roots of unity. I claim that $\varphi_3$ is the desired example.
We start by giving an explicit description of $C_3$. One easily calculates that the equations $x^2+y^2=1$ and $(x,y)^3 = (1,0)$ are equivalent to $x = 1-2y^2$ and $3y = 4y^3$. So we have an alternative description
$$ C_3 = \{ (x,y) : x = 1-2y^2, 3y = 4y^3 \} \cong \{ y : 3y = 4y^3 \}. $$
It is easy to see that $\varphi_3$ has trivial kernel: If $\varphi_3(x,y) = 1$, we also have $\varphi_3(x,-y) = \varphi_3(x,y)^{-1} = 1$, and hence $0 = \varphi_3(x,y) - \varphi_3(x,-y) = 2i y$. But then also $3y = 4y^3 = 0$, whence $y = 0$. Now $x=1-2y^2=1$ is trivial.
At this point we have seen that the affine group scheme $\mu_3$ has an affine subgroup $C_3$ isomorphic to $\{ y : 3y = 4y^3 \}$. It is clear that these group schemes are not isomorphic since $C_3$ collapses to the trivial group after base change to any ring with $2 = 0$, while $\mu_3$ stays non-trivial. So it remains to show that $\mu_3$ is the scheme theoretic image of $\varphi_3$
This is equivalent to show that the induced homomorphism on coordinate rings
$$ \varphi^* \colon R[T]/(T^3-1) \to R[Y]/(3Y - 4Y^3), \quad T \mapsto 1-2Y^2+iY $$
is injective. This can be done entirely without computations: Note that we already know that $\varphi^*$ becomes an isomorphism after tensoring with any $R$-algebra containing $1/2$. In particular, $\varphi^* \otimes_R Q(R)$ is an isomorphism, where $Q(R) = \mathbb{Q}[i]$ is the field of fractions of $R$. Since tensoring with the field of fractions only kills torsion elements, and since $R[T]/(T^3-1)$ is a free (hence torsion-free) $R$-module, the kernel of $\varphi^*$ must be trivial.
