In elementary physics, it is well-known that the Newton's law $$\vec{F}=m\vec{a}$$
is invariant under Galilean transformations. However, Galilean relativity is not introduced in details in ordinary textbooks on Quantum Mechanics and Classical Mechanics. Naively, one usually treats the Galilean group as a spacetime symmetry (i.e global and physical symmetry), just as Lorentzian symmetry in QFT.
In relativistic QFT, one usually applies the Noether's theorem to find conserved quantities associated with these global symmetries, such as energy, momentum and angular momentum etc, because Lorentz transformations are rigid.
It is, however, not the case in a generic Galilean transformation, which is shown below $$t^{\prime}=t+a,\quad\vec{r}^{\prime}=R\vec{r}+\vec{v}t+\vec{b},$$
where $a\in\mathbb{R}$, $\vec{v}$, $\vec{b}\in\mathbb{R}^{3}$, and $R\in O(3)$.
In the above equations, a pure Galilean boost clearly depends on time parameter. In other words, the transformation of a classical trajectory in the Lagrangian $L(\dot{\vec{r}})$ explicitly depends on $t$, and so this is a local (i.e gauge) transformation.
To be specific, I am talking about the following transformation of a trajectory of a classical particle (I am not talking about field theory) $$\vec{r}(t)\rightarrow\vec{r}(t)+\vec{v}t,$$
which clearly depends on the parameter $t$. To be even more specific, this transformation takes the form $$\vec{r}(t)\rightarrow\vec{r}(t)+\vec{f}(t),$$
where $\vec{f}(t)=\vec{v}t$ is a linear function of $t$. In other words, this transformation in Classical Mechanics is not rigid.
To avoid further irrelavent information in the comment section, please notice that the $c\rightarrow\infty$ limit of the Lorentz group is not the Galilei group. This was clarified by David Bar Moshe and Qmechanic ♦ here. Unlike the Lorentz group, the Galilei group has a non-trivial central extension, parameterized by the mass of the free Newtonian particle. When one lifts its projective representation, one ends up with the linear representation of its universal central extension.
However, unlike gauge transformations in QED which leaves the Lagrangian density invariant, a Galilean boost changes the Lagrangian of a non-relativistic free particle by a total derivative.
In QFT, there's a huge difference between a physical (global) symmetry and a gauge (local) redundancy. Physical symmetries are in irreducible unitary representations that map states into some other states, while gauge redundancies are in trivial representations that leave the quantum states invariant. So what is going on here with Galilean transformations? It's Eucliean subgroup is clearly physical, but its pure boost seems almost like a gauge transformation.