Are there Goldstone bosons in the spontaneous normal to superfluid transition?

The Goldstone boson for a superfluid is the phonon.

See Wikipedia:

A version of Goldstone's theorem also applies to nonrelativistic theories (and also relativistic theories with spontaneously broken spacetime symmetries, such as Lorentz symmetry or conformal symmetry, rotational, or translational invariance).

It essentially states that, for each spontaneously broken symmetry, there corresponds some quasiparticle with no energy gap—the nonrelativistic version of the mass gap. [...] However, two different spontaneously broken generators may now give rise to the same Nambu–Goldstone boson. For example, in a superfluid, both the U(1) particle number symmetry and Galilean symmetry are spontaneously broken. However, the phonon is the Goldstone boson for both.

And also this article:

An example of spontaneous symmetry breaking is the breaking of the global U(1)-symmetry in $^4$He and the appearance of superfluidity below a certain critical temperature. Associated with the breaking of the symmetry, there is a nonzero value of an order parameter and a condensate of particles in the zero-momentum state. When a global continuous symmetry is broken, the Goldstone theorem states that there is a gapless excitation for each generator that does not leave the ground state invariant. In the case of nonrelativistic Bose gases, one identifies the Goldstone mode with the phonons.

Other relevant sources:

• A. Schmitt, Introduction to Superfluidity
• K. Huang, Statistical Mechanics (chap. 13)