The group $A = {\rm AGL}(1,k)$ is a semidirect product $N \rtimes H$, where $N$ is isomorphic to the additive group of the field, and H to its multiplicative group, and the action of $H$ on $N$ is defined by field multiplication - so it is fixed-point-free.
If $k$ does not have characteristic 2,then we have $-1 \in H$ has order 2, and $H$ is its centralizer in $A$. All elements of order 2 are conjugate in $A$ and it follows that all complements of $N$ in $A$ are conjugate to $H$. The same is true if $K$ is finite. I am not sure whether it is necessarily true for infinite fields of characteristic 2.
Assuming that, since $N$ is clearly characteristic in $A$, any automorphism of $A$ must fix $N$ and we can assume, by multiplying it by an inner automorphism, that it fixes $H$. It is then easy to check that it must be induced by an automorphism of the field $K$. So ${\rm Aut}(A) = {\rm A \Gamma L}(1,k)$ and ${\rm Out}(A) \cong {\rm Aut}(k)$. So in the case $k = {\mathbb Q}$, there are no outer automorphisms.
Any nontrivial normal sub group of $A$ contains $N$, so an endomorphism that was not an automorphism would have $N$ in its kernel and would effectively be a homomorphism from $H$ to $A$. The image would be abelian so would either be a subgroup of $N$ or isomorphic to a subgroup of $H$ itself, and there could be many possibilities.
Added later: In addition, if the field $k$ has proper subfields $k'$ isomorphic to itself, then there could be iinjectiv but not surjective endomorphisms that map ${\rm AGL}(1,k)$ to ${\rm AGL}(1,k')$. That would not happen of course with $k = {\mathbb Q}$.
