First of all, even setting aside the issue with scalar automorphisms noted in comments, at the level of objects there is a "problem": the functor is *not* essentially surjective for unipotent groups (by a consideration of Ext$^1$-groups with their natural structure of *vector space* over the ground field). Probably nobody cares, so I won't get into the details (but if someone does care then hopefully someone else will summon the energy to write out an actual argument, since I don't feel like going down that road here).

Now let's focus on positive statements that someone might care about.
Let $K'/K$ be an extension of fields (any characteristic, but characteristic 0 is especially nice when $G^0$ is reductive because of the Remark below), and $G$ a smooth affine group over $K$. Assume $K$ is algebraically closed (but $K'$ is not assumed to be algebraically closed). Then we claim:

**Theorem**: *For any semisimple linear representation $\rho':G_{K'} \to {\rm{GL}}(V')$, there exists a semisimple linear representation $\rho:G \to {\rm{GL}}(V)$ unique up to isomorphism such that $\rho' \simeq \rho_{K'}$.*

Remark: In characteristic 0 the semisimplicity hypothesis on the representations automatically holds if $G$ has reductive identity component. There is a version of the Theorem in characteristic 0 that doesn't require $K$ to be algebraically closed if $G$ is split connected reductive, but that involves an entirely different ingredient than is used below, so I'll pass over it in silence.

To prove the Theorem, we begin with:

**Lemma** *The coefficients of the characteristic polynomial of $\rho'$ come from $K[G] \subset K'[G_{K'}]$.*

*Proof*. An element $f' \in K'[G]$ comes from $K[G]$ if and only if its restriction to a Zariski-dense subset of $G(K)$ belongs to $K$.
(Indeed, if we apply "spread out and specialize" to $f’$ then we get some $f \in K[G]$ such that $f_{K'}$ agrees with $f'$ on a Zariski-dense
subset of $G(K)$, but such a subset is also Zariski-dense in $G_{K'}$ (exercise!), so $f' = f_{K'}$ as desired.)

Let $B$ be a Borel $K$-subgroup of $G$, so the $G(K)$-conjugates of $B(K)$ cover $G(K)$ (since $K$ is algebraically closed). Hence, it suffices to show that the coefficients on all such $G(K)$-conjugates have values in $K$. Let $T$ be a maximal $K$-torus in $B$, so $B = T \ltimes U$ for $U := \mathscr{R}_u(B)$. Applying Lie-Kolchin to $B_{K'}$ acting on $V'$, those coefficients on any $b \in B(K)$ only depend on the $T$-component of $b$. Thus, since the characteristic polynomial is conjugation-invariant, we're reduced to studying these coefficients on points of $T(K)$. All weights of $T_{K'}$ are "$K$-rational" (since $T$ is a split $K$-torus, as $K$ is algebraically closed), so we win.

QED Lemma

Now if we apply "spreading out and specialization" followed by semisimplification, from $\rho'$ we get a semisimple
$$\rho: G \to {\rm{GL}}(V)$$
such that $\rho_{K'}$ has the same characteristic polynomial as $\rho'$ due to the Lemma. But $\rho_{K'}$ is semisimple because $\rho$ is semisimple with $K$ algebraically closed (i.e., $V$ is a semisimple representation of the abstract group $G(K)$ with $K$ algebraically closed, so it is "absolutely semisimple" and hence — by consideration of the endomorphism algebra — $V_{K'}$ is semisimple over $K'$ as a representation of the abstract group $G(K)$ and thus as a representation of the algebraic group $G_{K'}$ by Zariski-density of $G(K)$ in $G_{K'}$). Hence, $\rho_{K'}$ and $\rho'$ are semisimple representations of $G(K')$ with the same characteristic polynomial, so by Brauer-Nesbitt (which applies to semisimple representations of finite dimension for any abstract group at all) these representations are isomorphic. That isomorphism amounts to a single conjugation in the language of matrices, so it says that as algebraic representations they’re $K'$-isomorphic. This gives that $\rho'$ descends to a semisimple representations of $G$ over $K$ as desired.

QED Theorem

groupalong with the field, which makes the question less straightforward.. $\endgroup$ – Jim Humphreys Dec 23 '17 at 2:10notessentially surjective (whether or not we consider the category of $K$-reps with morphisms tensored with $L$). Indeed, consider a 1-dimensional rep of $L^2$ (viewed as 2-dimensional abelian unipotent) whose kernel is not defined over $K$. Then this rep is not isomorphic to any rep in the image of the functor (since isomorphisms preserves kernels). $\endgroup$ – YCor Dec 23 '17 at 3:48