Given $A$ and $H$ linear algebraic groups over a field $k$ and a homomorphism $\phi\colon H \rightarrow \mathrm{Aut}(A)$ we can form the semi-direct product $G = A \rtimes_{\phi} H$.
Suppose that $G$ acts on a scheme $X$. The splitting $H\rightarrow G$ gives the restriction map $K^{0}(G,X)\rightarrow K^{0}(H,X)$ in equivariant algebraic K-theory.
$\textbf{Question:}$ Is this map an isomorphism of groups if we assume that $A$ is unipotent?
If $X=\mathrm{Spec}(k)$ then $K^{0}(G,X)=R(G)$ is the group completion of $\mathrm{Rep}(G)$. I think that restriction $R(G)\rightarrow R(H)$ is an isomorphism because the only finite dimensional irreducible representation of $A$ is the trivial one, so $R(A)\simeq \mathbb{Z}$. The inverse is given by precomposition by the projection $\pi \colon G \rightarrow H$.
If $H$ is trivial then $G \simeq A$ is unipotent and we get that $K^{0}(G,X)\rightarrow K^{0}(X)$ is an isomorphism, according to Thomason.
In the setting of Chow rings an analogous statement (with rational coefficients) is Lemma 1.4.7 of Brokemper, "On the Chow ring of the stack of truncated Barsotti-Tate groups".
