Preamble
Consider the Lagrangian density for electrodynamics: $$L=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-A_\mu J^\mu\tag{1}$$ With the usual definition of $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$. This results in the Euler-Lagrange equations: $$\partial_\rho F^{\rho\mu}=J^\mu\tag{2}$$ From which we can determine that $$\partial_\mu J^\mu=0,\tag{3}$$ so the vector field can only be consistently coupled to a conserved current. Now let's say we want to fix the gauge via a Lagrange multiplier, using Coulomb gauge. Thus: $$L=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-A_\mu J^\mu+\lambda\partial_i A_i.\tag{4}$$ The Euler-Lagrange equations now are: $$\partial_i F^{i0}=J^0,\tag{5}$$ $$-\partial_0 F^{k0}+\partial_i F^{ik}-\partial_k \lambda=J^k,\tag{6}$$ $$\partial_i A_i=0.\tag{7}$$ We have Coulomb gauge, but it looks like the equations of motion have changed with the appearance of the Lagrange multiplier field $\lambda$ in equation (6).
In fact though, using the condition that $\partial_\mu J^\mu=0$, one can show, by adding the time derivative of (5) to the divergence of (6), that: $$\partial_i \partial_i \lambda =0.\tag{8}$$ So that (at least with appropriate boundary conditions?) $\lambda$ is constant and the equations of motion are not changed after all.
Question
My question is, does this hold generally? Given a gauge theory (e.g., $SU(N)$ Yang-Mills), if one adds the gauge condition as a constraint with a Lagrange multiplier, are the resulting Euler-Lagrange equations always equivalent to the original ones, just supplemented with the gauge condition?
