Is there any conclusion about a group, if the group has unique element of order $n>1$? If a group $G$ has an unique element of order $n>1$, then which of the following is true:


*

*Order of $G$ is $n$.

*Order of $Z(G)$ is greater than $n$.

*$Z(G)=G$

*$G=S_2$


(I've seen that (1) can not be true. Because order of $\mathbb{Z}_5$ is $5$, but there are $4$ elements of order $5$ in $\mathbb{Z}_5$.) 
Firstly, we know that in any group the number of elements of order $n$ is $\phi (n)$ or its multiple, if $n$ divides the order of the group.
Now, $\phi (n)=1$ gives $n=1$, $n=2$.
But, according to the question, n>1.
So $n=2$ only.
i.e. it's given that the group contain an unique element of order $2$.
But which option is correct between $2, 3, 4$?
 A: Clearly $o(n) = o(n^{-1})$ so this means there must be exactly one element of order $2$. If you mean this for all $n$, then this means the group must just be $\Bbb Z/2\Bbb Z$ which is isomorphic to $S_2$. So $1$, $3$, $4$ all hold.
On the other hand if you just mean this to be true for some $n$, I must think you're crazy since again we see that $n$ is $2$, and since the group is characteristic, it means that for any $x\in G$ if the unique order $2$ element is $z$ then $xzx^{-1}=z\implies z\in Z(G)$, so $G$ has a non-trivial center. Clearly the Center need not be non-trivial, as you can do $S_2\times H$ with $H$ an odd, trivial-centered group and the center of a product of groups is the product of their centers. In any case, the first one cannot work since $\Bbb Z/4\Bbb Z$ has a unique element of order $2$ but the order is $4$ and the previous example of a direct product shows the order need not exceed $n$ and need not be equal to $|G|$. And if so, then none of them are necessarily true.
Also, for your example, $\Bbb Z/5\Bbb Z$ does not have a unique element of any order (except, of course, the identity), so it does not fit the context of the problem.
