Suppose I have a Lagrangian density $\mathcal{L}(\phi^\mu,\sigma)$ depending on vector fields $\phi^\mu$ and their derivatives and a scalar field $\sigma$ and its derivatives. If I make a gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$ does the field $\sigma$ transform? I've seen notes claiming $\sigma \rightarrow \sigma + \alpha$ but just wanted to make sure.
Edit: For more detail - I was looking at a problem that said the Lagrangian $\mathcal{L}(\phi^\mu)=-\frac{1}{2}(\partial_\mu\phi^\nu)^2+\frac{1}{2}(\partial_\mu\phi^\mu)^2+\frac{m^2}{2}(\phi_\mu\phi^\mu)^2$ is not invariant with respect to the gauge transformation $\phi^\mu\rightarrow \phi^\mu+\partial^\mu\alpha$ and asked to introduce a new scalar field $\sigma$ and find a new interacting Lagrangian $\mathcal{L}'(\phi^\mu,\sigma)=\mathcal{L}(\phi^\mu)+\tilde{\mathcal{L}}(\phi^\mu,\sigma)$ which is gauge invariant under the given transformation and satisfies $\mathcal{L}'(\phi^\mu,0)=\mathcal{L}(\phi^\mu)$.
I've found a Lagrangian with these properties but it assumes $\sigma \rightarrow \sigma + \alpha$ and I'm not sure if this is correct.