Quantum symmetries: $S$ or $Z$?

Question

Let $I$ be the action of some QFT (gauge-fixed and including all the necessary counter-terms); $S$ the associated scattering-matrix; and $Z$ the partition function (in the form of, say, a path integral). There are three notions of symmetries that are typically discussed,

1. Action symmetries, that is, transformations of the form $\phi\to\phi'$ that leave $I$ invariant,
2. $S$-matrix symmetries, that is, operators that (super)commute with $S$, and
3. Quantum symmetries, that is, transformations of the form $\phi\to\phi'$ that leave the volume form $\mathrm e^{I[\phi]}\mathrm d\phi$ invariant.

It is a well-known phenomenon that the symmetries of $I$ need not agree with those of neither $S$ nor $Z$ (e.g., anomalies, SSB, etc.). What is not-so-clear is whether $S$ symmetries and $Z$ symmetries are equivalent. What I want to know is whether

For each symmetry of $S$ there is a symmetry of $Z$ and vice-versa

or a counter-example. If the equivalence is actually true, I would like to have a more-or-less precise statement, in the form of a theorem (to the usual level of rigour in physics textbooks).

No cheating please. Counter-examples are only valid for "real-life" QFT's (e.g. in a free theory everything commutes with $S$ but not everything leaves $Z$ invariant; this is not a valid counter-example because it is completely trivial). No TQFT's either. Thanks.

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^{Someone mentioned in the comments the effective action $\Gamma[\phi]$, which is defined as the Legendre transform of $\log Z$. I didn't want to bring this object into the picture, because I wanted to leave it to the discretion of the rest of users whether to mention this object or not. In principle, I don't need answers to analyse the symmetries of this object, but they may if they believe it could be useful to do so. In any case, let me stress that $\Gamma$ is not the same object as $I$.}

Answer 1

People sometimes talk about on-shell symmetries: symmetries of the equations of motion or the S-matrix, which do not hold off-shell (i.e. at the level of the action, path integral, correlators, etc). Electric-magnetic duality transformations $$ \begin{aligned} F &\rightarrow \cos(\theta)\,F - \sin(\theta)\,\star F \\ \star F &\rightarrow \sin(\theta)\,F + \cos(\theta)\,\star F \end{aligned} $$ are an example of this. These transformations leave the equations of motion and the S-matrix invariant (with a corresponding selection rule), so they would qualify as a symmetry.

However, this symmetry is usually broken by non-perturbative effects (e.g. monopoles). In addition, it is well known that dualities are not symmetries of the full theory but instead a better understood as a change in our description of the physics. For instance duality transformations relate the partition function of the theory at different values of the coupling (see this paper for an example).

Answer 2

I think that a symmetry in Z implies perturbatively a symmetry in S since the perturbative S-matrix can be computed from the effective action (which is directly related to Z) in a straightforward way that leaves no room for breaking a symmetry.

Whether the converse holds is questionable as it is not even clear whether the S-matrix determines the action.