There is no obstruction to descent from potential automorphy to automorphy lie in some Galois. Conjecturally, potential automorphy should be equivalent to automorphy.
To see why, let us take as our primary object that can be automorphic or not, Galois representation instead of varieties or motives (to fix ideas). Specifically, we consider continuous irreducible representations $r:G_K \rightarrow Gl_n(\bar {\bf Q}_p)$, where $K$ is a number field, $G_K$ its absolute Galois group, and $p$ prime number. Let us recall the following "folklore" conjecture
Conjecture (Fontaine-Mazur + Langlands). A Galois representation $r$ of $G_K$ as above is automorphic if and only if $r$ is unramified at almost all places of $K$ and potentially semi-stable at every place of $K$ dividing $p$.
Here automorphic means that $r$ has the same $L$-factors at almost all primes that a cuspidal automorphic algebraic representation $\pi$ of $GL_n(\bf{A}_K)$.
Now fix $K'$ a finite extension of $F$it is almost obvious that the conditions after the "if and only if are satisfied for $r$ if and only if they are satisfied for $r_{|G_{K'}}$. Thus automorphy for $r$ should be the same as automorphy for $r_{|G_{K'}}$. In author words, automorphy is conjecturally the same as potential automorphy.
Now why is it not possible to prove on the nose that if $r_{G_K'}$ is automorphic, that is associated with an a automorphic representation $\pi'$ of $GL_n({\bf A}_{K'})$, then $r$ itself is automorphic? Because that would mean to "descend" $\pi'$ into s suitable automorphic representation $\pi$ of $GL_n(\bf{A}_K)$, and this question, called "automorphic descent", is very hard. It is part of the extremely hard Langlands functoriality, and has been solved so far only in the case $K'/K$ solvable (by Langlands for $n=2$, Arthur-Clozel for general $n$).