If we replace the single axiom "Emergence" in the axiomatic system $T$ presented at posting [What is the set theory synonymous with this order-set theory], by the following schema. Would the resulting theory still be synonymous with $\sf PA$? If no, is there a kind of restricted synonymy notion that it would have with $\sf PA$?
$ \textbf {Emergence: } i=0,1,2,\dots; \ n=i-1, \dots, -1;\\ \max_i x < \max_i y \land (\bigwedge_n \max_n x = \max_n y ) \to x < y $
$ \max_{-1}x = \varnothing \\\max_0 x = \min y: \forall m \in x \, (m \leq y) \\ \max_{n+1} x = P_x(\max_n x); \text { if } n > -1$
Where $P_x$ is the predecessor relation on $x$. This is: $$ a=P_x(b) \iff a \in x \land b \in x \land a < b \land \not \exists c \in x: a < c < b$$
Where, "$\max_i x < \max_i y$" is defined as:
$ \exists b \, \bigl( b=\max_i y \land \forall a \, (a= \max_i x \to a < b) \bigr)$
While, " $\max_i x = \max_i y$" is defined as:
$\exists a \exists b \, (a=\max_i x \land b=\max_i y \land a=b)$
Best Answer
Take a nonstandard model of $\mathsf{PA}$, and let $\omega$ be a nonstandard natural number. The Ackermann interpretation gives a model of $T$. Now, we take the sets $\{n | n \leq \omega\}$ and $\{n | 1 \leq n \leq \omega\}$ (which are adjacent in the ordering), and swap them. The result does not satisfy Emergence, but it does satisfy every (standard) axiom in the Emergence schema, because $\max_n$ of either set is just $\omega-n$ (there is no axiom about $``\max_\omega"$ because $\omega$ is nonstandard). The axioms other than Emergence are all still true, so this is a model of $T$ + Emergence schema + $\neg$ Emergence. Call this model $M$.
$T$ without Emergence interprets $\mathsf{PA}$ by my earlier answer, so from $M$ we have a model $\mathbb{N}_M$ of $\mathsf{PA}$. $\mathsf{PA}$ interprets $T$, so from $\mathbb{N}_M$ we have a model $N$ of $T$. $M$ and $N$ are not elementarily equivalent, because $N$ satisfies Emergence and $M$ doesn't. Now, to prove that $T$ with the Emergence schema is not bi-interpretable with $\mathsf{PA}$, we can apply Emil Jeřábek's argument to $M$ and $N$, with some minor changes: $M$ and $N$ are not actually well-ordered, but they think they are. So $M$ thinks $M^F$ is the standard model of $\mathsf{PA}$. The embedding of $N$ in $M$ preserves $<$, so $M$ also thinks that $N$ is well-ordered, and therefore that $N^F$ is also the standard model of $\mathsf{PA}$, so $M^F \simeq N^F$ (and the argument from there is the same).
So this theory is not synonymous with $\mathsf{PA}$.