Suppose we are in Euclidean 3-space with coordinates $x$ and a scalar function $\phi(x)$ defined on it, and consider the group of rotations $SO(3)$ for simplicity. Take a rotation matrix $R \in SO(3)$; then the usual explanation for the difference between active and passive transformation is:
Active: If we picture the function $\phi(x)$ as having a "bump" somewhere, an active transformation moves this bump around. We now have a new function $\phi'$ of the same coordinates $x$, given by $\phi'(x) = \phi(R^{-1}x)$.
Passive: We leave the bump fixed and instead make $R$ rotate the coordinate axes to get a new set of coordinates $x'$ on the same physical space. In terms of the new coordinates, the field is expressed as a function $\phi'(x') = \phi(Rx')$ or $\phi'(x) = \phi(Rx)$ if you prefer, since the name of the coordinates doesn't make any difference.
This makes perfect sense but there's a problem: the passive transformation is not a representation of the group of rotations. If we are to think of transformations as an action of $SO(3)$ (or the Lorentz group or whatever your favorite group is) on the space of functions, we can only use active transformations, since only they are actually a group action. Does this imply that if we study fields as representations of some group we are restricting ourselves to active transformations only? Shouldn't the active and passive viewpoints be equivalent?
Edit: the passive transformation is not a group representation because if we define $(\rho_R \phi)(x) = \phi(Rx)$, we get $\rho_{R_1} \rho_{R_2} = \rho_{R_2 R_1}$ instead of the other way around.