While thinking about how to introduce $0^\sharp$ to students, the following weak analogue occured to me:
Say that a transitive structure $\mathcal{A}=(A;\in, ...)$ - that is, a transitive set possibly equipped with finitely many additional relations - is good for a club of countable ordinals $C$ iff ($A\cap \mathsf{Ord}=\omega_1$ and) the length-$\omega_1$-sequence $(C;\in)$ is $\mathcal{A}$-indiscernible but the Skolem hull of $C$ in $\mathcal{A}$ is all of $A$. A good club is a club for which there exists a good transitive structure. If we strengthen the hypotheses on $\mathcal{A}$ to include mild closure properties then the existence of a good club is equivalent to $0^\sharp$, but even with no additional hypotheses goodness seems nontrivial. (For example, any "reasonably thin" club is indiscernible for the structure $(\omega_1;\in)$ but fails to generate that structure EDIT: this is wrong!.)
My question is whether the statement "There exists a good club" is equivalent, over $\mathsf{ZFC}$, to "$0^\sharp$ exists." In case it isn't, what is its strength over $\mathsf{ZFC}$? (Embarrassingly, it's not immediately clear to me that it isn't outright $\mathsf{ZFC}$-provable.)
As has come up a few times on MO, surprisingly little is definable in ordinals with just their order structure. (In some sense this shouldn't be surprising, all linear orders are NIP, and so are fairly model-theoretically tame.) The structure $(\omega_1,\in)$ has QE in the language that contains $\in$, constant for $0$, and functions $f_\alpha(x) = x+\alpha$ for each $\alpha < \omega^\omega$.
Since we have QE to a language only containing at most binary predicates and at most unary functions, a sequence is indiscernible if and only if it is indiscernible for pairs. From this it's easy to see that the sequence $\{\omega^\omega\cdot \alpha :\alpha <\omega_1\}$ is indiscernible, and by the QE characterization, this has all of $\omega_1$ as its Skolem hull. Therefore this is a good club.