Why are some group representations much easier to compute discrete logarithm for?
Well, the group being isomorphic doesn't imply that the isomorphism is efficiently computable. If $G \simeq H$ via $\phi : G \rightarrow H$ and $\phi$ is computable, then indeed, DLOG is no harder in $G$ than in $H$ because you can transform the instance.

In the case you mentioned, we have $G = \mathbb Z/(p-1) \mathbb Z$ and $H = (\mathbb Z/p\mathbb Z)^\times$. A efficient morphism $\phi: G \rightarrow H$ is known, it called `fast exponentiation'. So DLOG is no harder in $G$ than it is in $H$. But computing a reversed morphism $\psi: H \rightarrow G$ seems harder...