The answer is: 2, 4, 8 and 3, 6, 12, 24. I can prove it, but I suspect there is a simpler proof.
First, notice that $|\mathbb{Z}^{\times}_{n}| = \varphi(n)$, and hence that $\varphi(n)$ must be a power of 2. If $p^2$ divides $n$ then $p$ divides $\varphi(n)$, so if $p > 3$ then $p$ divides $n$ at most once. Suppose $p |n$ exactly once. Then by the Chinese Remainder Theorem,
$$\mathbb{Z}^{\times}_{n} \cong \mathbb{Z}^{\times}_{n/p}\times \mathbb{Z}^{\times}_{p}$$
but $\mathbb{Z}^{\times}_{p} \cong C_{p-1}$ has an element of order $p-1 > 2$. Therefore the only options are $3 \cdot 2^n$ and $2^n$. Manual checking shows $n=1,2,3$ works in the first case and $n=0, 1, 2, 3$ in the second case. Mod $2^k$ for $k \ge 4$ we have $3^2 = 9$, and mod $3 \cdot 2^k$ for $k \ge 4$ we have $5^2 = 25$.
That's short enough now, but I'm skipping a lot of details. The nearby exercises were all pretty easy, but this took me a while. Am I missing an easier way?
