This question is closely connected to the following paper and to a prior question posted to Mathoverflow titled "Can Cantor's theorem be proved in Paraemter Free Zermelo",

https://wwwmath.uni-muenster.de/u/rds/ZFC_without_parameters.pdf

That paper provided an answer to a question about whether $\text{ZFC}^o$ is equivalent to $\text{ZFC}$, where $\text{ZFC}^o$ is the theory axiomatized by axioms of Extensionality, Foundation, Pairing, Union, Power, Infinity and a Parameter free version of Replacement and Separation denoted by $\text{Repl}^o$, $\text{Aus}^o$ respectively; where the axioms of Pairing, Union and Power are written in full, i.e. one can prove existence the sets $\{a,b\}$; $U(a)$; $P(a)$ for any sets a,b using those three axioms and Extensionality without using an instance of schema of specification.

In the Wikipedia, axioms of $\text{ZFC}$ are written in a different manner, see the following link:

https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory

The axioms of Pairing, Union, Power and Replacement only asserts existence of a non specific set that contains sets a,b ; elements of elements of a; subsets of a, replacements of elements of a respectively; one needs to use Specification in order to prove existence of the sets $\{a,b\}$; $U(a)$; $P(a)$, $F(a)$

Now if by $\text{eZFC}^o$, denoting $\text{"extreme Parameter Free ZFC"}$, it is meant $\text{ZFC}$ axiomatically presented as how it is mentioned in the Wikipeida but with Specification and Replacement axioms replaced by parameter free versions of them, i.e. by $\text{Repl}^o$ and $\text{Aus}^o$ with these two schemata being formalized exactly as in the above-mentioned article (see: Postscript) for full exposition of $\text {eZFC}^o$. Then apparantly the proof presented in the above article fails, because it clearly depends on specifying pairs, unions, powers and replacment sets without the need to use Specification, and thus in some manner it allows the use of Parameters external to Specification scheme. Along the same line $\text{eZC}^o$ denotes $\text{"extreme Parameter Free Zermelo"}$, and this is $\text{eZFC}^o$ minus Replacment.

Given that, the questions are:

Is $\text{eZFC}^o$ equivalent to $\text{ZFC}$?

Is $\text{eZC}^o$ equivalent to $\text{ZC}$?

If both are false then is Cantor's theorem provable in $\text{ZFC}^o$ or $\text{ZC}^o$?

PS: $\text{eZFC}$ is formulated in the first order language of set theory with the following axioms:

Extensionality: $\forall A,B (\forall X (X \in A \iff X \in B) \iff \forall Y (A \in Y \iff B \in Y))$

Foundation: $\forall A (\exists X \in A \implies \exists B \in A (\not \exists C \in B (C \in A)))$

Specification: if $\phi$ is a formula in which only symbol $X$ occurs free, then

$\forall A \exists B \forall X (X \in B \iff X\in A \wedge \phi)$

is an axiom.

Pairing: $\forall A,B \exists C (A \in C \wedge B\in C)$

Union: $\forall A \exists B \forall X,Y (X \in Y \wedge Y \in A \implies X \in B)$

Power: $\forall A \exists B \forall X (\forall Y \in X (Y \in A) \implies X \in B)$

Replacement: If $\phi(Y,Z)$ is a formula in which only symbols $Y,Z$ occur free, then:

$\forall A \exists B \forall X (\exists Y \in A (\forall Z (\phi(Y,Z) \iff Z=X)) \implies X \in B)$

is an axiom.

Infinity: $\exists X (\varnothing \in X \wedge \forall Y \in X (\{Y\} \in X))$

$\text {+\-}$

$AC$