Let $\mathscr{T}$ be the (countable) collection of all theorems provable in ZFC. Define an equivalence relation on $\mathscr{T}$ by $\phi\sim\psi$ iff $(\phi \iff \psi)$. In other words, two theorems are equivalent iff they are logically equivalent.
Then we can say $\mathscr{T}$ is the disjoint, countable union of equivalence classes of (equivalent) theorems, call them $E_i$. By the axiom of choice, there exists a set, $\mathscr{E}$, containing precisely one element (theorem) from each equivalence class: $\mathscr{E} = \{\phi_i : i = 1, 2, …\}$, $\phi_i \in E_i$.
Now, two theorems $\phi$ and $\psi$ are contained in $\mathscr{E}$ iff they belong to disjoint equivalence classes. In other words, $\phi, \psi \in \mathscr{E}$ iff $\neg (\phi \iff \psi)$ iff $((\phi \wedge \neg\psi) \vee (\neg\phi \wedge \psi))$, in particular $\phi$ and $\psi$ cannot both be true, contradicting the fact that they are theorems.
Best Answer
You're tacitly assuming that $\mathscr{E}$ contains more than a single element.
But suppose that $\varphi$ and $\psi$ are theorems of $\mathsf{ZFC}$. Then in particular, in every single model $M$ of $\mathsf{ZFC}$, $\varphi$ and $\psi$ are true. As such, we find that $M\vDash (\varphi\leftrightarrow \psi)$ for every model $M$, so by Completeness we see that $\varphi\leftrightarrow \psi$ is a theorem of $\mathsf{ZFC}$. This implies that $\mathscr{E}$ consists of a single element, because there was only a single equivalence class to start with.
(EDIT: Here I'm assuming you mean "logically equivalent relative to $\mathsf{ZFC}$". Refer to Asaf's answer to understand the distinction.)