Why is this functor corepresentable? Consider the functor from the category of commutative algebras to groups that sends every algebra to the group of strict automorphisms of its additive formal group law. Can you tell me, why this functor is corepresentable? Can you also tell me what the corresponding Hopf algebra is and the structure maps of this Hopf algebra?
I would be very happy if somebody could give me an answer.
 A: To specify a strict automorphism of the additive formal group is to specify a power series
$$g(x) =  x + b_1 x^2 + \dots$$
with the property that
$$g(x+y) = g(x) + g(y),$$
i.e. that $g$ is additive. This translates to a collection of polynomial conditions on the $\{b_i\}$, which means that the object corepresenting the functor $R \mapsto \{\mathrm{strict \ automorphisms \ of \ } \hat{\mathbb{G}_a} \}$ is the quotient of $\mathbb{Z}[b_1, b_2, \dots ]$ by the aforementioned relations. 
When tensored with $\mathbb{Q}$, this ring is $\mathbb{Q}$ itself: that is, the additive formal group admits no nontrivial strict automorphisms over $\mathbb{Q}$. (In fact, the groupoid of formal group laws over a $\mathbb{Q}$-algebra and strict isomorphisms is discrete, and equivalent to the one-point groupoid.) This amounts to saying that there are no additive power series over $\mathbb{Q}$. (These notes of Neil Strickland have a discussion of formal group laws over $\mathbb{Q}$, and much more. The more general statement for not-necessarily-commutative formal group laws in higher dimension is that they are equivalent to Lie algebras, so once you know what happens on the tangent space you know the whole map.) 
Over a $\mathbb{F}_p$-algebra, there can be automorphisms: for instance, $x + x^2$ in characteristic $2$. In this case (over $\mathbb{F}_2$), the ring classifying strict isomorphisms of the additive formal group is the dual Steenrod algebra (by a theorem of Milnor; see this question for a discussion). Over $\mathbb{F}_p$, the ring is isomorphic to dual of the algebra of Steenrod reduced powers (the dual Steenrod algebra is only graded commutative). 
For the global structure, note that the ring is naturally graded with $b_i$ in degree $i$ (this grading comes from the action of the multiplicative group given by conjugation with the automorphism $\lambda t$, $\lambda$ invertible). Each graded piece is a finitely generated abelian group, and you can work out what they are from the structure rationally and the structure at each prime: the  zeroth one is $\mathbb{Z}$, the first one is $\mathbb{Z}/2$, etc.  
