You have $3$ possibilities:
$$x^3=x^5 \Rightarrow x^2=e \,.$$
In this case since $x^{15}=e$ it follows that $x=e$, which is not possible (since you only get one value in your set).
$$x^5=x^9 \Rightarrow x^4=e \,.$$
Again, this implies that $x=e$, not possible.
$$x^3=x^9 \Rightarrow x^6=e \,.$$
Thus, $x^3=x^{\operatorname{gcd}(6,15)}=e$. This means that $x$ must have order $1$ or $3$, but again $x=e$ is not possible.
Thus, $x$ is an element of order $3$ in your group, and from there it is easy: $x^{13n}=x^{13m} \Leftrightarrow 3|13(m-n) \Leftrightarrow 3|m-n$... So how many distinct values do you get?
You are right, the group $D_n$ contains exactly one cyclic subgroup of order $n$ when $n > 2$. To prove this, you can use the fact that every nonidentity element outside $C$ has order $2$.
Best Answer
The easiest one is in $\mathbb{Z}_4$. The element 2 generates the subgroup $\{2,0\}$, and it's the only one, since all the other elements have order 4 or 1.
More generally, every $\mathbb{Z}_{2d}$, has this property, since the element $d$ generates $\{0,d\}$
Given $G$ finite group with odd order, then $\mathbb{Z}_{2}\times G$ is also ok.
switching to theory, we have that $a$ must be the only element in $G$ with order 2, so it is invariant by conjugation. It means that $H=\{0,a\}$ is a normal subgroup of $G$. If $G$ is a finite group, and $H$ is a $2$- Sylow, then $G=H\times G'$.