Definition field of isogeny between abelian varieties Let $K$ be a number field. Let $A$ and $B$ be abelian varieties over $K$. Assume that $A$ and $B$ are isogenous over $\bar{K}$, the algebraic closure of $K$. We further assume that the endomorphism ring of $A_{\bar{K}}$ is $\mathbb{Z}$. The question is:
Does there exist a quadratic extension $L$ over $K$ such that $A$ and $B$ are isogenous over $L$?
 A: Yes. The group of isogenies form a locally free module of rank $1$ over the endomorphism ring of $A$, hence are generated by a single isogeny of minimal degree $k$. So every isogeny is that isogeny composed with an endomorphism of $k$, so has degree $n^{2g} k$ for $n \in \mathbb Z$. Hence there are two isogenies with a given degree. The Galois action on the set of isogenies preserves degree, so the stabilizer of any isogeny is an index $2$ subgroup.
Here I'm just thinking of an isogeny as a special kind of subvariety of $A \times B$, from which it is clear that Galois invariance implies definability. 
A: Yes. This is easy to see in terms of Galois cohomology. Let's work in the category $\mathcal{A} \otimes \mathbb{Q}$ of "abelian varieties up to isogeny". Then for an object $A$ of $\mathcal{A} \otimes \mathbb{Q}$, we have 
$$\operatorname{Aut}(A_{\overline{K}}) = \mathbb{Q}^\times = \left\langle - 1 \right\rangle \oplus \bigoplus_{p \textrm{ prime}} \mathbb{Z},$$
so a twist of $A$ (that is, an abelian variety that is $\overline{K}$-isogenous to  $A$) defines a class in
$$
\operatorname{H}^1(K,\operatorname{Aut}(A_{\overline{K}})) =\operatorname{H}^1(K,\left\langle - 1 \right\rangle \oplus \bigoplus_{p \textrm{ prime}} \mathbb{Z}) = \operatorname{H}^1(K,\left\langle - 1 \right\rangle) = K^{\times}/K^{\times 2},
$$
with the isomorphisms all being canonical in $K$; moreover, two twists give the same class iff they are isomorphic in $\mathcal{A}\otimes \mathbb{Q}$. (Here I have used that $\operatorname{H}^1(K,\mathbb{Z})=0$ in the next-to-last step, and the Kummer sequence in the last.)
Therefore, if $B$ is as in your question, say corresponding to $t \in K^{\times}/K^{\times 2}$, then $A_L$ is isogenous  to $B_L$ iff $\sqrt{t} \in L$.
