= Solution
For fields $\Phi_a$ with a first-derivative <Lagrangian density>, suppose an infinitesimal global transformation at fixed coordinates is $\delta\Phi_a=\alpha\Delta_a$ and $\delta\mathcal L=\alpha\partial_\mu K^\mu$, with constant infinitesimal $\alpha$. <Noether theorem> gives the <on shell> <conserved current>
$$
\boxed{J^\mu=\sum_a\frac{\partial\mathcal L}{\partial(\partial_\mu\Phi_a)}\Delta_a-K^\mu,\qquad \partial_\mu J^\mu=0.}
$$
This follows by varying the Lagrangian, integrating the terms with $\partial_\mu\Delta_a$ by parts, and setting the Euler-Lagrange expressions to zero. If spatial boundary flux vanishes, $\int d^3x\,J^0$ is a conserved <Noether charge>. Spacetime symmetries are covered by including the induced field variation at fixed coordinates and the corresponding total derivative $K^\mu$.
An example of a theory with a Lorentz <four-vector> is the source-free <Maxwell field> with <electromagnetic four-potential> $A_\mu$. Under an Abelian <gauge transformation>,
$$
A_\mu\longmapsto A_\mu+\partial_\mu\chi,\qquad
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.
$$
The change in $F_{\mu\nu}$ is $\partial_\mu\partial_\nu\chi-\partial_\nu\partial_\mu\chi=0$, so the <Maxwell Lagrangian> is <gauge-invariant>. For $A_\mu=(\Phi,-\mathbf A)$ define $E_i=F_{0i}$ and $F_{ij}=-\epsilon_{ijk}B_k$. With signature $+---$,
$$
F_{\mu\nu}F^{\mu\nu}=2(\mathbf B^2-\mathbf E^2),\qquad
\mathcal L=\frac12(\mathbf E^2-\mathbf B^2).
$$
The minus sign in $-F^2/4$ gives the positive electric kinetic term and positive physical field energy; reversing it would reverse the <Hamiltonian>'s sign for the physical modes. The normalization $1/4$ accounts for antisymmetry and yields the conventional equations. Varying $A_\nu$ gives $\partial_\mu F^{\mu\nu}=0$.
For time translation at fixed coordinates take $\Delta_\nu=\partial_0A_\nu$. Since $\alpha$ is constant, $\delta F_{\mu\nu}=\alpha\partial_0F_{\mu\nu}$, and hence $\delta\mathcal L=\alpha\partial_0\mathcal L$. Thus $K^\mu=\delta^\mu{}_0\mathcal L$. The momentum derivative is
$$
\frac{\partial\mathcal L}{\partial(\partial_\mu A_\nu)}=-F^{\mu\nu}.
$$
<Noether theorem> therefore gives the <canonical stress-energy tensor>'s time-translation current
$$
J_{\rm can}^\mu=T_{\rm can}^\mu{}_0
=-F^{\mu\nu}\partial_0A_\nu-\delta^\mu{}_0\mathcal L,
\qquad
\boxed{H_{\rm can}=\int d^3x\left[\frac12(\mathbf E^2+\mathbf B^2)+\mathbf E\cdot\nabla A_0\right].}
$$
Here $\dot A_i=E_i+\partial_iA_0$ was used. The <canonical momentum> of $A_0$ vanishes, while that of the lower spatial component $A_i$ is $E_i$: $A_0$ acts as a multiplier for <Gauss law>, not a propagating degree of freedom. The density above is not manifestly <gauge-invariant>, but its charge agrees with the physical energy on the constraint surface and with appropriate <boundary conditions>.
The <gauge-covariant time translation> instead has $\Delta_\nu=\partial_0A_\nu-\partial_\nu A_0=F_{0\nu}$. It is time translation combined with a <gauge transformation> of parameter $\chi=-\alpha A_0$. Directly,
$$
\delta F_{\mu\nu}=\alpha(\partial_\mu F_{0\nu}-\partial_\nu F_{0\mu})
=\alpha\partial_0F_{\mu\nu},
$$
using the definition of $F$, equivalently its <Bianchi identity>. The total derivative in $\delta\mathcal L$ is unchanged. The improved current is
$$
J_{\rm imp}^\mu=-F^{\mu\nu}F_{0\nu}-\delta^\mu{}_0\mathcal L
=\Theta^\mu{}_0,
\qquad
\Theta^\mu{}_{\nu}=-F^{\mu\rho}F_{\nu\rho}+\frac14\delta^\mu{}_{\nu}F_{\rho\sigma}F^{\rho\sigma}.
$$
It is the <gauge-invariant Maxwell stress-energy tensor>. In particular,
$$
\boxed{u=J_{\rm imp}^0=\frac12(\mathbf E^2+\mathbf B^2),\qquad
H=\int d^3x\,u.}
$$
The spatial current is the <Poynting vector> $\mathbf E\times\mathbf B$, so conservation gives $\partial_tu+\nabla\cdot(\mathbf E\times\mathbf B)=0$. This is the familiar positive local <electromagnetic field> energy density, expressed entirely in measurable fields.
To compare the currents,
$$
J_{\rm can}^\mu-J_{\rm imp}^\mu=-F^{\mu\nu}\partial_\nu A_0
=-\partial_\nu(A_0F^{\mu\nu})+A_0\partial_\nu F^{\mu\nu}.
$$
The last term vanishes on the source-free <Maxwell equations>, leaving a <stress-energy tensor improvement> generated by a <stress-energy superpotential>. At the charge level,
$$
H_{\rm can}-H=\int d^3x\,\mathbf E\cdot\nabla A_0
=\oint A_0\mathbf E\cdot d\mathbf S-\int d^3x\,A_0\nabla\cdot\mathbf E=0
$$
when the surface term vanishes and <Gauss law> holds. Thus the symmetries yield the same conserved energy under these conditions while the second supplies a gauge-invariant local density. Boundary terms would have to be retained if those <boundary conditions> were not imposed.
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