It is best to never let anyone get away with saying that all physical theories are generally covariant. Gravity is, but that's about it. If you want to consider passive co-ordinate transformations, you are just performing a substitution in the integral and
\begin{align}
S = \int d^dx \mathcal{L}[\phi(x)] = \int d^dx^\prime \left | \frac{dx^\prime}{dx} \right |^{-d} \mathcal{L}[\phi(x^\prime)]
\end{align}
are two equivalent ways of writing the action. We usually choose the former but this is just notation. There is a chance to learn something useful about the theory if we also consider active transformations. In this case,
\begin{equation}
S^\prime = \int d^dx^\prime \left | \frac{dx^\prime}{dx} \right |^{-d} \mathcal{L}[\phi^\prime(x^\prime)]
\end{equation}
is a different action because the $\phi^\prime(x^\prime)$ are different functions. How they might be expressible in terms of $\phi(x)$ again (i.e. what representation they furnish) is up to us or whomever handed us the theory. If they conspire to make $S^\prime = S$ then the particular transformation we used will be a symmetry of the theory.
For obvious reasons, fields are often defined to be in irreducible representations of the Lorentz group so that we know how to vary the action under a Lorentz transformation. And, unless we are doing something strange, the Lagrangian will also contract all indices so that we indeed get zero. For scale transformations, the situation is a bit different. Representations are again known ahead of time since the engineering dimension is easy to read off. But now the variation of an action like
\begin{equation}
S = \int d^dx \frac{1}{2} \partial_\mu \phi \partial_\mu \phi + \frac{1}{2} m^2 \phi^2
\end{equation}
yields a term proportional to $m^2$ which is an intrinsic scale of the theory. This tells us that even though we can use centimeters instead of meters, there is no conformal symmetry. Making your ruler 100 times as long won't give you the same predictions anymore unless you also change the theory to one where the mass is 100 times as small.
As the previous answer points out, the need to consider how fields transform is paramount. However, specifics about the infinite dimensional enhancement of conformal symmetry that exists in two dimensions are tangential. The point is that, whether or not you work with fields known to transform as tensors, only very special theories are given by an action which also does.
