Show the automorphism group of $\mathbb{Z}_{49}$ is isomorphic to $\mathbb{Z}_{42}$ I'd like to see how $Aut(Z_{49})\cong Z_{42}$. I know that $Aut(Z_{49})\cong (\mathbb{Z}/42\mathbb{Z})^x$ where $Z_k$ is the cyclic group order k, but I fail to see how either of these is isomorphic to $Z_{42}$, or even that it is cyclic for that matter.

You have that $\mathrm{Aut}(\mathbb{Z}_{49}) \simeq \mathbb{Z}_{49}^{\times}$, not $\mathbb{Z}_{42}^{\times}$.  Now you want to show that $\mathbb{Z}_{49}^{\times} \simeq \mathbb{Z}_{42}$.
This group does not have too many elements, so just check to find a generator.  Once you get more than 21 distinct powers of an element, you are guaranteed that element is a generator.
