If your system consists of (a) a uniform electric field along direction $\hat{\mathbf{n}}$, or (b) a uniform magnetic field along direction $\hat{\mathbf{n}}$, then its symmetries are
- translation orthogonal to $\hat{\mathbf{n}}$,
- translation parallel to $\hat{\mathbf{n}}$, and
- rotations about $\hat{\mathbf{n}}$.
Some of these transformations are equivalent to a gauge transformation. However, gauge transformations are still symmetries of the system, as they respect the physics. More specifically, the motion of a charged particle in those fields is preserved by the symmetries listed above, and it is the motion that matters.
This is somewhat clearer if one spells it out explicitly. Consider a particle of charge $q$ under the action of the electric and magnetic fields $\mathbf E=E_0\hat{\mathbf{z}}$ and $\mathbf B=B_0\hat{\mathbf{z}}$. The hamiltonian for the system can always be written as
$$
H=\frac1{2m}\left(\mathbf p-q\mathbf A\right)^2+\varphi,
$$
where the potentials $\mathbf A=\mathbf A(x,y,t)$ and $\varphi=\varphi(z,t)$ must satisfy
$$
\nabla\times\mathbf A=\mathbf B=B_0\hat{\mathbf{z}}
\quad\text{and}\quad
-\frac{\partial\mathbf A}{\partial t}-\nabla \varphi=\mathbf E=E_0\hat{\mathbf{z}}.
$$
Sample potentials that satisfy this are $\varphi_0=E_0z$ and $\mathbf A_0=\tfrac12 B_0(x\hat{\mathbf{y}}-y\hat{\mathbf{x}})$, and all other acceptable potentials can be obtained from these via
$$
\mathbf A=\mathbf A_0+\nabla\chi
\quad\text{and}\quad
\varphi=\varphi_0-\frac{\partial\chi}{\partial t}
$$
for a suitable function $\chi=\chi(x,y,z,t)$, which has no influence at all on the physics.
Suppose now that you transform the system with the transformation above: You translate by a vector $\mathbf a$ and you rotate about $\hat{\mathbf z}$ with a rotation matrix $R$, so you take $\mathbf r \mapsto \mathbf r'=R(\mathbf r+\mathbf a)$ and $\mathbf p\mapsto \mathbf p'=R\mathbf p$. In terms of the new variables, the old hamiltonian is
\begin{align}
H
&=
\frac1{2m}\left(\mathbf p-q\mathbf A(\mathbf r,t)\right)^2+\varphi(\mathbf r,t)
\\&=
\frac1{2m}\left(R^{-1}\mathbf p'-q\mathbf A(R^{-1}\mathbf r'-\mathbf a,t)\right)^2+\varphi(R^{-1}\mathbf r'-\mathbf a,t)
\\&=
\frac1{2m}\left(\mathbf p'-qR\mathbf A(R^{-1}\mathbf r'-\mathbf a,t)\right)^2+\varphi(R^{-1}\mathbf r'-\mathbf a,t),
\end{align}
because $R$ is orthogonal and preserves norms, so $(R\mathbf u)^2=\mathbf u^2$. This can be written in the same way as before
$$
H\mapsto H'=\frac1{2m}\left(\mathbf p'-q\mathbf A'(\mathbf r',t)\right)^2+\varphi'(\mathbf r',t),
$$
for the transformed potentials
$$
\mathbf A'(\mathbf r',t)=R\mathbf A(R^{-1}\mathbf r'-\mathbf a,t)
\quad\text{and}\quad
\varphi'(\mathbf r',t)=\varphi(R^{-1}\mathbf r'-\mathbf a,t).
$$
We now use the fact that the forms of the old potential are constrained to be gauge transforms of a standard set, so
\begin{align}
\mathbf A'(\mathbf r',t)&=R\mathbf A_0(R^{-1}\mathbf r'-\mathbf a,t)+R\nabla\chi(R^{-1}\mathbf r'-\mathbf a,t)
\quad\text{and}\quad\\
\varphi'(\mathbf r',t)&=\varphi_0(R^{-1}\mathbf r'-\mathbf a,t)-\frac{\partial\chi}{\partial t}(R^{-1}\mathbf r'-\mathbf a,t)
,\end{align}
and the fact that we know how the standard set transforms: writing the rotation matrix explicitly as
$$
R=\begin{pmatrix}\cos\theta&\sin\theta & 0\\ -\sin(\theta)&\cos \theta & 0\\0&0&1\end{pmatrix}
$$
we can transform the potentials to
\begin{align}
R\mathbf A_0(R^{-1}\mathbf r'-\mathbf a,t)
&=
\frac{B_0}{2}
\begin{pmatrix}\cos\theta&\sin\theta & 0\\ -\sin(\theta)&\cos \theta & 0\\0&0&1\end{pmatrix}
\begin{pmatrix}
-\sin\theta x'-\cos\theta y'+a_y
\\
\cos\theta x'-\sin\theta y'-a_x
\\ 0
\end{pmatrix}
\\&=
\frac{B_0}{2}
\begin{pmatrix}
-y'+\cos\theta a_y-\sin\theta a_x
\\
x'-\sin\theta a_y-\cos\theta a_x
\\0
\end{pmatrix}
\\&=
\mathbf A_0(\mathbf r',t)+\mathbf b
\end{align}
and
\begin{align}
\varphi_0(R^{-1}\mathbf r'-\mathbf a,t)=E_0(z'-a_z)=\varphi_0(z',t)-\phi,
\end{align}
where $\mathbf b$ and $\phi$ are constant.
This transformation is now explicitly in the form of a gauge transformation, with $\chi=\mathbf b·\mathbf r-\phi t$. As I explained before, gauge transformations do not change the physics, so this completes the proof.
