Are symmetries spontaneously broken in a CFT? I have a vague intuition – most likely incorrect – that all (internal) flavor symmetries in a  CFT are always spontaneously broken. (Perhaps under some mild assumptions, such as unitarity and that the conformal symmetry itself is unbroken).
Is this actually provable? Or are there any known counterexamples (i.e., a CFT with a faithful global symmetry that is unbroken)? Are there counterexamples for both continuous and discrete symmetries?
I mostly care about $d>2$ since Coleman–Mermin–Wagner won't allow continuous symmetries to break in 2d. But of course if there is anything known about 2d specifically I would also like to hear it.
 A: If you have a symmetry which rotates between vacua, it is always possible to take operators which transform trivially under it and build up an algebra from them on top of one arbitrarily chosen vacuum. So you are looking for a construction which yields a CFT of the small type instead of the big type.
How about the $O(2)$ model? To study it with the bootstrap, we assume that getting a non-zero two-point function for the order parameter only requires one vacuum. I.e. the normalization condition looks like $\left < 0 | \phi^i(0) \phi^j(\infty) | 0 \right > = \delta^{ij}$ instead of $\left < 0 | \phi^i(0) \phi^j(\infty) | 0^\prime \right > = \delta^{ij}$. This results in striking agreement with experimental critical exponents.
Could it be that single vacuum $O(2)$ symmetric CFTs with $\Delta_\phi = 0.509$ actually don't exist? Because the observed second order phase transition is governed by a CFT whose structure is rich enough that you lose the order parameter when you try to restrict the number of vacua to one? Yes because numerics do not give a proof. But a wrong assumption sending such a clear signal that we are on the right track would be shocking.
