= Solution
Take $\mathbf x'=\mathbf x-\mathbf u t$, with the same time coordinate in both frames. The <normal fluid> transforms as $\mathbf v_n'(\mathbf x',t)=\mathbf v_n(\mathbf x,t)-\mathbf u$. The field transformation is the unit-modulus Schrödinger boost
$$
\boxed{\psi'(\mathbf x',t)=e^{-i\mathbf u\cdot\mathbf x+i|\mathbf u|^2t/2}\psi(\mathbf x,t),
\qquad\mathbf A=-i\mathbf u,\quad B=-\frac i2|\mathbf u|^2.}
$$
Writing the phase as $F=-i\mathbf u\cdot\mathbf x'-i|\mathbf u|^2t/2$ makes differentiation at fixed primed position transparent:
$$
\nabla'\psi'=e^F(\nabla\psi-i\mathbf u\psi),\qquad
\partial_t'\psi'=e^F\left(\psi_t+\mathbf u\cdot\nabla\psi-\frac i2u^2\psi\right).
$$
It follows that both $-i\partial_t'\psi'$ and $\nabla'^2\psi'/2$ gain the same terms $e^F[-i\mathbf u\cdot\nabla\psi-u^2\psi/2]$. The density is unchanged, $\rho'=|\psi'|^2=\rho$, so the nonlinear local term has the same form.
The remaining <material derivative> transforms as
$$
(\partial_t'+\mathbf v_n'\cdot\nabla')\rho'
=[\partial_t+\mathbf u\cdot\nabla+(\mathbf v_n-\mathbf u)\cdot\nabla]\rho
=(\partial_t+\mathbf v_n\cdot\nabla)\rho.
$$
Thus the density-damping term acquires only the common phase factor, proving invariance of the complete equation under a <Galilean boost>. Keeping the normal <velocity> unchanged in both frames would fail this check; the reference flow must transform along with the condensate.
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