Model-companion of $T_\sigma$ Let $T$ be a first-order $L$-theory, and let $T^*$ be the model-companion of $T$. Also, let $T_\sigma$ be the extension of $T$ with a generic automorphism $\sigma$, and $(T_\sigma)^*$ be the model-companion of $T_\sigma$.
Question. Are $(T_\sigma)^*$ and $(T^*)_\sigma$ the same?
 A: In $(T^*)_\sigma$, the behavior of the automorphism $\sigma$ is completely unconstrained, while in $(T_\sigma)^*$, the automorphism $\sigma$ behaves generically. In particular, if $T$ is already model-complete, then $T = T^*$, so $(T^*)_\sigma = T_\sigma$, while $(T_\sigma)^*$ is its model companion. 
For example, if you are familiar with the theory $\text{ACFA}$ of algebraically closed fields with a generic automorphism, we have if $T = \text{ACF}$, then $(T^*)_\sigma = \text{ACF}_\sigma$ and $(T_\sigma)^* = \text{ACFA}$. Or if $T$ is the theory of an infinite set in the empty language, then $(T^*)_\sigma$ is the theory of an infinite set with an arbitrary permutation, while $(T_\sigma)^*$ is the theory of an infinite set with a permutation $\sigma$ which has infinitely many fixed points and infinitely many cycles of each finite length. 
Also, for most theories $T$ (no longer assuming model completeness), we have that $(T_\sigma)^*$ entails the sentence $\exists x\,(\sigma(x) \neq x)$ (to prove this for a theory $T$, you just have to show that any model of $T_\sigma$ can be extended to a larger model in which $\sigma$ is not the identity), while $(T^*)_\sigma$ does not (since $(M,\text{id}_M)\models (T^*)_\sigma$ whenever $M\models T^*$). 
