Show $\displaystyle \mathbb{Z}_{14}^* \cong \mathbb{Z}_{18}^* $.
$\displaystyle
\mathbb{Z}_{14} ^ \times
$ and $\displaystyle
\mathbb{Z}_{18} ^ \times
$ have the same order and both are cyclic, hence they are isomorphic.
(Remember that if $\displaystyle p>2$ is prime then $\displaystyle
\mathbb{Z}_{2p^k } ^ \times
$ is cyclic for all $\displaystyle k\in \mathbb{Z}^+$, to see this, read about primitive roots)