Hom(G_{a}, G_{m}) is not representable.

Let R be a ring of characteristic zero. I claim that Hom(G_{a}, G_{m})(Spec R) is {Nilpotent elements of R}. Intuitively, all homs are of the form x -> e^{nx} with n nilpotent.

More precisely, the schemes underlying G_{a} and G_{m} are
Spec R[x] and Spec R[y, y^{-1}] respectively. Any hom of schemes is of the form y -> \sum f_i x^i for some f_i in R. The condition that this be a hom of groups says that
\sum f_k (x_1+x_2)^k = (\sum f_i x_1^i) (\sum f_j x_2^j). Expanding this, f_{i+j}/(i+j)! = f_i/i! f_j/j!. So every hom is of the form f_i = n^i/i!, and n must by nilpotent so that the sum will be finite.

Now, let's see that this isn't representable. For any positive integer k, let R_k = C[t]/t^k. The map x -> e^{tx} is in Hom(G_{a}, G_{m})(Spec R_k) for every k. However, if R is the inverse limit of the R_k, there is no corresponding map in Hom(G_{a}, G_{m})(Spec R). So the functor is not representable.