Symmetry for dipole conservation in field theory In article The Fracton Gauge Principle complex scalar field is considered.
There's statement, that for conservation of charge one needs usual U(1) global symmetry:
$$
\phi \to e^{i\alpha}\phi \Rightarrow Q =\int d^Dx \rho
$$
For conservation of dipole moment:
$$
\phi \to e^{i\vec x \cdot \vec \lambda}\phi \Rightarrow Q^i =\int d^Dx x^i\rho
$$
Why for dipole moment we need such symmetry?
 A: For a symmetry
$$ \phi \rightarrow \mathrm{e}^{\mathrm{i}\lambda(x)}\phi, $$
Noether's theorem gives you a conserved four-current:
$$ J^\mu = \mathrm{\lambda}\frac{\partial \mathcal L}{\partial\frac{\partial \phi}{\partial x_\mu}}\phi $$
which satisfies the continuity equation:
$$ \partial_\mu J^\mu = 0 \Leftrightarrow \frac{\partial J^0}{\partial t} + \nabla \cdot\mathbf{J} = 0. $$
If there are no current sources or sinks, then $\nabla \cdot\mathbf{J} = 0$ and the quantity $J^0$ is conserved:
$$ \frac{\partial J^0}{\partial t} = 0.$$
Let's take the Dirac Lagrangian as an example, which gives you
$$ J^\mu = \lambda \bar\psi \gamma^\mu\psi.$$

*

*With $\lambda = \alpha$, then the conserved quantity $J^0$ is:
$$ Q = \int\mathrm{d}^3\mathbf{r} \, J^0 =  \int\mathrm{d}^3\mathbf{r} \, \alpha |\psi|^2,$$
where $\bar\psi = \psi^\dagger$ allowing to take the absolute mod squared, $|\psi|^2$ being the (normalised) probability density. $\alpha$ is then the electrical charge of the particle under consideration.


*With $\lambda(x)= x^i \lambda_i$, then the conserved quantity $J^0$ is:
$$ Q = \int\mathrm{d}^3\mathbf{r} \, J^0 =  \int\mathrm{d}^3\mathbf{r} \, x^i \lambda_i |\psi|^2 = \lambda_i \int\mathrm{d}^3\mathbf{r} \, x^i |\psi|^2 ,$$
where $\int\mathrm{d}^3\mathbf{r} \, x^i |\psi|^2$ is the definition of the $p^i$ dipole moment, $|\psi|^2$ being the charge distribution.
Now, the conserved quantity is actually $\boldsymbol{\lambda}\cdot \mathbf{p}$, but I guess you can take $\boldsymbol{\lambda}$ to point along $x$, or $y$, or $z$ and each of these would give you a conserved $p_x, p_y$, or $p_z$. Because $\boldsymbol{\lambda}$ is arbitrary?
A: I think the right way to think about it is to first ask what the symmetry
$$
\phi \longrightarrow e^{i\theta} \phi
$$
means. The actual Unitary operator acting on the many-body Hilbert space that corresponds to this transformation is
$$
U(\theta) = \exp\left( i\theta \sum_r \phi^\dagger_r \phi_r \right)
$$
It is easy to see that $\phi$ transforms as shown in the first equation upon the adjoint action of $U$. So you see that the the $U(1)$ symmetry is generated by the total particle number operator. So, a system that has this symmetry conserves particle number and vice versa.
Now we consider a system with conservation of dipole moment. Dipole moment is given by the vector operator $\sum_r \boldsymbol{r}\phi^\dagger_r \phi_r $. Therefore, conservation of dipole moment actually refers to three independent conservation laws, whose corresponding unitary operator will be parametrised by a 3-vector $\boldsymbol{\lambda}$ as
$$
U(\boldsymbol{\lambda}) = \exp\left( i\boldsymbol{\lambda} \cdot \sum_r \boldsymbol{r}\phi^\dagger_r \phi_r \right)
$$
Now one can ask how does this act on the particle annihilation operator at a particular site $\phi_{r_0}$. We can simply look at the adjoint action
\begin{equation}
\begin{split}
\phi_{r_0} \longrightarrow U^\dagger\phi_{r_0} U &= \exp\left( -i\boldsymbol{\lambda} \cdot \sum_r \boldsymbol{r}\phi^\dagger_r \phi_r \right) \phi_{r_0} \exp\left( i\boldsymbol{\lambda} \cdot \sum_r \boldsymbol{r}\phi^\dagger_r \phi_r \right) \\
&=\exp(i\boldsymbol{\lambda} \cdot \boldsymbol{r_0})\phi_{r_0}
\end{split}
\end{equation}
Thus we see that what you stated is indeed the transformation that is generated by the total dipole moment operator, so it is the corresponding symmetry to it. Note that we can easily generalise this argument to the continuum.
