Let $M$ be a model of ${\sf ZFC}$. A satisfaction class $S$ for $M$ is subset of $M$'s ordered pairs which satisfies in $M$ the standard Tarskian compositional axioms. E.g.:

$M\vDash \forall \phi, \psi\in\mathcal L_\in\forall a:\omega \to V(\langle\phi\wedge \psi, a\rangle\in S\leftrightarrow \langle\phi, a\rangle\in S \wedge \langle \psi, a\rangle\in S)$

**Question:** Is there a model $M$ of ${\sf ZFC}$ with two satisfaction classes $S$ and $S'$ such that for some formula $\phi$ and assignment function $a$ in $M$, $\{x: \langle\phi, a[0/x]\rangle\in S\} = S'$ (where $a[0/x]$ is the assignment function according to $M$ which is just like $a$ except that it assigns $x$ to $0$)?

Obviously, $M$ will have to be non-standard, as will $\phi$. Similarly, by Tarski's theorem on the undefinability of truth, $S$ and $S'$ can't agree on pairs of formulas and assignment functions. And as Joel notes below, a simple induction using a predicate for $S$ would show that $S$ and $S'$ do agree on formulas and assignment functions. So $M$ will not be a model of ${\sf ZFC}$ with Separation extended to a predicate for $S$.  

I'm primarily interested in models of ${\sf ZFC}$, but I'd also be interested in models of ${\sf PA}$.