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 ${\sf PA}$. 



Can one satisfaction class code another?
In case of Peano Arithmetic the answer is yes (emphatically yes), if I understand your question correctly. This follows from Theorem 3.3 in Smith's "Nonstandard Definability" APAL 42 (1989) pp. 21-43 which says that for any recursively saturated countable model $M$ of PA and any set $A \subseteq M$ the set $A$ may be presented in a form $\{x \in M \mid (\phi, x) \in S \}$ for some satisfaction class $S$ in $M$ iff the expansion $(M,A)$ is recursively saturated (I'm slightly confused with the $0$ in your notation and I know I am a bit sloppy with writing of a pair (a formula, an element) being in a satisfaction class rather than (a formula, an assignment), but I hope this is what you looked for).


Now take any $A$ in the model $M$ which is a recursively saturated satisfaction class (being a satisfaction class can be finitely axiomatised and being recursively saturated is axiomatised by a scheme, so by resplendency of countable recursively saturated models and the fact that the theory of satisfaction class without induction (or replacement in your case) is conservative over PA there exists such an $A$ in $M$).


So we conclude that: in any countable recursively saturated model $M$ of PA there exists a class $S$ which defines another satisfaction class or even a chain of such classes.


I haven't really checked whether this proof carries over to ZFC, but I would be extremely surprised if it were not the case. So I'm 99 % sure the answer to your question is very strong yes.


