Let $T$ be a totally transcedental theory and $M$ a model of $T$ with $A \subset M$. Let $p(x) \in S(A)$ (a complete type with parameters in $A$) and let $q(x)$ be a nonforking extension of $p(x)$ with parameters in $M$. I am trying to understand how $q(x)$ behaves under an $A$-automorphism of $M$.
I know that given another non-forking extension $s(x)$ of $p(x)$ (to all of $M$), there exists an $A$-automorphism of $M$ , say $f$ such that $f(q(x)) = s(x)$ (This is result prop 2.33 in http://www.math.uiuc.edu/People/pillay/lecturenotes.stability.pdf). But what I am wondering is given an arbitrary $A$-automorphism $f$, is $f(p(x))$ going to be a nonforking extension of $p(x)$? That is, do the $A$-automorphisms permute the nonforking extensions of $p(x)$?
Yes, the $A$-automorphisms permute the nonforking extensions of $p(x)$.
The reason depends on which definition of nonforking you're using. Since you linked to Pillay's lecture notes, I'll assume that you make the definition $q$ doesn't fork over $A$ if and only if $q$ is definable almost over $A$.
Let $q$ be a nonforking extension of $p$, and let $f$ be an $A$-automorphism. Now the $\phi(x,y)$ part of $q$ is definable by $\psi(y,a)$, with $a\in \text{acl}^{eq}(A)$, so the $\phi(x,y)$ part of $f(q)$ is definable by $\psi(y,f(a))$. Any automorphism fixing $A$ fixes (setwise, not pointwise) $\text{acl}^{eq}(A)$, so $f(a)\in\text{acl}^{eq}(A)$, and $f(q)$ is definable almost over $A$. Finally, $q\restriction A = p$, since the formulas with parameters in $A$ are fixed by $f$. So $q$ is a nonforking extension of $p$.