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Consider the covariant derivative $D_{\mu}=\partial_{\mu}+ieA_{\mu}$ of scalar QED.

$D_{\mu}\phi$ is invariant under the simultaneous phase rotation $\phi \rightarrow e^{i\Lambda}\phi$ of the field $\phi$ and the gauge transformation $A_{\mu}\rightarrow A_{\mu}+\frac{1}{e}(\partial_{\mu}\Lambda)$ of the vector potential $A_{\mu}$.

Are the phase rotation and the gauge transformation are related in any way? Does the gauge transformation of the vector potential necessarily lead to the phase rotation of the field?

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    $\begingroup$ There's a mistake in nomenclature. $D_\mu\phi$ isn't gauge invariant, it's gauge covariant. $(D_\mu\phi)^\dagger D_\mu\phi$ is gauge invariant. $\endgroup$
    – Sean E. Lake
    Commented Nov 3, 2016 at 0:14
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    $\begingroup$ What gauge covariant means is that $D_\mu \phi$, as a whole, transforms the same way $\phi$ does. That is: $\phi \rightarrow \operatorname{e}^{i\Lambda} \phi$ and $D_\mu\phi \rightarrow \operatorname{e}^{i\Lambda} D_\mu\phi$. $\endgroup$
    – Sean E. Lake
    Commented Nov 3, 2016 at 0:26

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A gauge transformation is defined to be the simultaneous transformation \begin{align} A_\mu & \mapsto A_\mu + e^{-1} \partial_\mu A \\ \phi & \mapsto \mathrm{e}^{\mathrm{i}\Lambda}. \end{align} It is not that the transformation of the gauge field "necessarily" leads to the transformation of the charged field, but only under this joint transformation everything is properly covariant. More abstractly, we conceive of both transformations to be different representations/realizations of the same abstract gauge algebra, consisting of functions $\Lambda : \mathbb{R}^4\to\mathbb{R}$. $\Lambda$ acts in one way on gauge fields and in another on charged fields, it's perfectly analogous to how the rotation group acts differently on scalars, vectors and tensors, yet "a rotation" is transforming all scalars, vectors and tensors and not just the vectors.

In a way, the covariance of $D_\mu\phi$ is the whole reason we invent the gauge field because $\partial_\mu \phi$ would not be covariant under the transformation of just $\phi$. It's less that the transformation of $\phi$ derives from that of $A$ and most that that of $A$ derives from that of $\phi$, at least if we take the viewpoint that the local symmetry is the fundamental object here. If we want to build a theory with such a local symmetry dependening on a spacetime function $\Lambda(x)$, we need to modify the derivative in order to build an invariant Lagrangian. So the first thing to try (perhaps motivated by the addition of the Christoffel symbols to the ordinary derivative in GR, which are effectively also only a type of gauge field) is to modify $\partial_\mu$ as $D_\mu := \partial_\mu + A_\mu$ for some $A_\mu$.

Now you examine $D_\mu (\mathrm{e}^{\mathrm{i}\Lambda}\phi)$, which we want to be equal to $\mathrm{e}^{\mathrm{i}\Lambda}D_\mu\phi$, and just look at the extra terms. After staring long enough at them, you will realize that those are exactly the terms that get cancelled if we let $A_\mu$ transform as we usually do, that is, the transformation of $A_\mu$ is engineered such that the covariant derivative is truly covariant.

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