Let G=A⋊B, where A and B are abelian, and of coprime order. It seems, from my computations (and correct me if I'm wrong), that Z^{1}(C_{p},C_{q}) is trivial, for p and q different primes. Meaning that the automorphisms of G, if A=C_{q}, and B=C_{p}, that preserve A, and preserve the cosets G/A, are all trivial. How far can we extend this? Would it be true in general that Z^{1}(A,B) is trivial, with the above assumptions (that A and B are abelian and of coprime order)? If not, under what assumptions is it trivial? And when can we say about it if it's not trivial?

The automorphisms of this extension are basically the same as group cohomology $H^1(B; A)$, so I will focus on that first. So we want to show that this must be zero given that B and A are finite abelian and of coprime order. This is the same as the twisted cohomology of the classifying space BB. Now we can look at the bundle $p:EB \to BB$. This is a covering space with fiber the discrete space B. Now because the fiber is discrete we have a wrong way transfer map in twisted cohomology: $$p_{!}: H^1(EB; A) \to H^1(BB; A)$$ where the first group is twisted cohomology in the pulledback local coefficient system. As with all transfers we have that $$p_{!} \circ p^*: H^1(B; A) \to H^1(B; A)$$ is multiplication by the order of the fiber, i.e. $B$. Since the orders of A and B are coprime this is an isomorphism. But since $EG \simeq pt$ is contractible, this map factors through the zero group and hence $H^1(B; A) = 0$. Any proof that the cohomology of a group is torsion for the order of the group (there are more concrete ones then the above) will yield the same result that $H^1(B; A) = 0$. There are many ways to prove this (as the comments point out), the above is just my favorite. So what is the difference between the $H^1(B;A)$ and the isomorphisms of the extension G? Well as you pointed out the isomorphisms of G (which restrict to the identity on A and the quotient B) are the same as the bar resolution coycycles Z^1(B;A). So we have an exact sequence, $$A= C^0(B;A) \to Z^1(B; A) \to H^1(B;A) \to 0$$ but as we saw, the term $H^1(B;A) = 0$. So we must compute the boundaries. You have one such potential homomorphism for each element of $A$, although different elements might give rise to the same automorphism of $G$. They are of the form: $$b \mapsto a  b \cdot a$$ where $a \in A$ is fixed. If the action of B on A is trivial, then these vanish, but in general they can be nonzero. My favorite example is the quaterion group which we view as $$\mathbb{Z}/4 \to Q_8 \to \mathbb{Z}/2$$ with the $\mathbb{Z}/4$ the group $( 1, i, 1, i )$. An element $x \in A = \mathbb{Z}/4$ induces the homomorphism $$y \mapsto x  y \cdot x$$ which sends the nontrivial element of $\mathbb{Z}/2$ to $2x$. In particular it is nontrivial for a generator of A. This corresponds to the isomoprhism of $Q_8$ which sends $i$ to $i$ and $j$ to $j$. 


You might be considering a special case of the SchurZassenhaus theorem. If A is a normal Hallsubgroup of G, then A has a complement B, and all complements B are conjugate under the action of G. This is more properly H^1(B,A) rather than Z^1(B,A). For Z^1(B,A) to be trivial, B^1(B,A) must be trivial, but B^1(B,A) is basically [B,A], which could be nonzero. For instance if G is nonabelian of order 6, then A is cyclic and normal of order 3, B is cyclic of order 2, and B acts as inversion on A. Then for B^1(B,A) should be isomorphic to A, that is have order 3. Basically, A has 3 complements in G, so B^1(B,A) should have three elements, but all are conjugate, so H^1(B,A)=0. I suppose there are lots of definitions of C^1(B,A), so maybe you've chosen one where B^1(B,A)=0, but I think the standard choice when looking at semidirect products ("Crossed homomorphisms"), will not have B^1(B,A) trivial. 

