The following scheme can be understood as a form of replacement. Axiomatizing $\sf ZF$ with it instead of the usual replacement schema renders it immune to removal of extensionality; see here.

In an article by Gitman/Hamkins/Johnstone, it was shown that the usual version of replacement is in some sense problematic if power set axiom is removed, while at the same time having the well-ordering principle.

Would the following version of replacement work as collection does if powerset is removed while having the well-ordering principle?

**Replacement$^*$:** if $\varphi$ is a formula in which $B$ is not free, then:

$$\forall A \exists B \forall y \, (y \in B \leftrightarrow \exists x \in A: \forall z (z \in y \leftrightarrow \varphi(x,z)))$$

If we have extensionality, then this is:

$$\forall A \exists B \forall y \, (y \in B \leftrightarrow \exists x \in A: y= \{z \mid \varphi(x,z)\})$$