$\mathbb Z_n^*$ has a unique subgroup of order $2$ then it is cyclic for $n >2$ If $\mathbb Z_n^*$ has a unique subgroup of order $2$  then it is cyclic, where $n>2$. That is, we have to show $n= 4, p^k,2p^k, p $ is odd prime.
I don't know if it is correct statement or not. But I suppose it is correct. If it is true please provide some hints to prove, otherwise please give some counter-example.

$Hint:$ If an abelian group $G$ can be written as $A \times B \times C$, with $A$ and $B$ of even order, then $G$ has at least two elements of order $2$.
Apply this to the decomposition
$$
(\mathbb{Z}/n\mathbb{Z})^\times\cong (\mathbb{Z}/{p_1^{k_1}}\mathbb{Z})^\times \times (\mathbb{Z}/{p_2^{k_2}}\mathbb{Z})^\times  \times (\mathbb{Z}/{p_3^{k_3}}\mathbb{Z})^\times \times \cdots
$$
given by the Chinese remainder theorem when $n=p_1^{k_1}p_2^{k_2}p_3^{k_3}\cdots$.
