Is this statement equivalent to AC?

Let $$Frege(X)$$ be the set of all equivalence classes of subsets of $$X$$ under equivalence relation bijection. formally:

$$Frege(X)= \{\{Y \subseteq X: |Y|=|Z|\}: Z \subseteq X \}$$

Where $$||$$ is cardinality function defined after Scott's.

My question is about the following statement:

$$\forall X: |Frege(X)| \leq |X|+1$$

Is it equivalent to the axiom of choice over the rest of axioms of $$\sf ZF$$?

More precisely can $$\sf ZF$$ plus the above statement prove the axiom of choice?

Afternote: To spell the above without invoking Scott's cardinality just replace $$|Y|=|Z|$$ by $$\exists f(f: Y \to Z \land f \text{ is a bijection})$$, and replace $$|Frege(X)| \leq |X| + 1$$ by $$\exists g (g: Frege(X) \to X \cup \{X\} \land g \text{ is an injection})$$

• I think invoking Scott cardinality just makes this messier: you're just asking about the principle "For all $X$ there is an injection from $\{$subsets of $X$ modulo bijectability$\}$ to $X\sqcup\{*\}$," right? Jun 28, 2020 at 23:48
• @NoahSchweber, yes definitely you can speak about them in terms in injections, etc... but it burns down to the same thing. Jun 29, 2020 at 0:41
• Yes, I only mentioned that since some readers might not be familiar with the notion of Scott cardinality. Jun 29, 2020 at 1:18
• OK, I've edited the post. Thanks! Jun 29, 2020 at 9:40