Show that $\mathbb{Z}^*_{15} \cong \mathbb{Z}^*_{16}.$

Since $\vert (\Bbb Z / n\Bbb Z)^\times\vert = \phi(n)$, we see that both groups have order $8$. They are clearly abelian, so they must be isomorphic either to $$\Bbb Z/8\Bbb Z,$$ $$\Bbb Z /4\Bbb Z \times \Bbb Z / 2 \Bbb Z$$ or $$\Bbb Z / 2 \Bbb Z \times \Bbb Z / 2 \Bbb Z \times \Bbb Z / 2 \Bbb Z.$$

Computing the orders of the elements shows that for both groups there are none of order $8$ while not all are of order $2$, so we are done.

There is a "smarter" method if you have some more results available. The group of units $U(pq)$ for two distinct odd primes $p$ and $q$ are isomorphic to $U(p)\times U(q)$, i.e., $$ U(15)\cong U(3)\times U(5)\cong C_2\times C_4. $$ And we have $$ U(2^n)\cong C_2\times C_{2^{n-2}} $$ in general, for all $n\ge 2$. Hence $U(16)\cong C_2\times C_4\cong U(15)$.

References:

Is my proof that $U_{pq}$ is not cyclic if $p$ and $q$ are distinct odd primes correct?

How to prove that $U_{2^n}$ is isomorphic as group to $\mathbb Z_2 \times \mathbb Z_{2^{n-2}}$ for $n \ge 3$?