Gauge transformation of Lagrangian 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.
 A: When you introduce a new field to make the Lagrangian gauge invariant, then you are at liberty to choose the transformation behaviour of the new field such that the Lagrangian becomes gauge invariant.
If $\sigma\mapsto\sigma+\alpha$ leads to an invariant Lagrangian, then you are free to choose $\sigma$ as a field transforming such. In the situation you presented, there's really not more to it.
Note though, that you have still not specified what the gauge group is, and if $\alpha$ is gauge Lie algebra valued rather than a real number, then you will have to let $\sigma$ take values in the Lie algebra as well rather than being just real valued (although it will still transform as a scalar under the Lorentz group).
A: 
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.

If I interpret what you wrote in the usual sense, then $\phi^\mu$ and $\sigma$ are independent field variables. Thus changing the $\phi^\mu$ in independent of changing the $\sigma$. What it looks like you are trying to say/understand is that the change $\phi^\mu\to\phi^\mu+\partial^\mu\alpha$ is equivalent to $\sigma\to\sigma+\alpha$.
Let me give you an example. Suppose that
$$
\mathcal{L}=x-y
$$
then the replacement
$$
x\to x+z
$$
is equivalent to the replacement
$$
y\to y-z
$$
because both have the same overall effect on $\mathcal{L}$.
