Find $\text{Aut}\left(\Bbb{Z}_{15}\right)$. Use the Fundamental Theorem of Abelian Groups to express this group as an external direct product of cyclic groups of prime power order.
For this would it just be as simple as $U(5) \oplus U(3)$ due to $3$ and $5$ being the only factors of $15$ in $\Bbb{Z}_{15}$? I feel as though I am definitely missing a key piece since this seems to be easy.
On
The automorphism group of a cyclic group is its group of units, in this case $\Bbb Z_{15}^×$.
The group of units functor respects products. Thus, since by the Chinese remainder theorem, $\Bbb Z_{15}\cong\Bbb Z_3×\Bbb Z_5$, we get $\Bbb Z_3^××\Bbb Z_5^×\cong \Bbb Z_2×\Bbb Z_4$.
(In the last step, I used that for any prime $p$, we have $\Bbb Z_p^×\cong\Bbb Z_{p-1}$.)
Yes, that is indeed one valid way to do it, you are using the chinese remainder theorem.
But to conclude you need to express $U(5)$ and $U(3)$, remember that they are isomorphic to the multiplicative group of $\mathbb Z_n$, which are cyclic for primes.
So the decomposition is $\mathbb Z_4\times \mathbb Z_2$