Solution: Since the order of the group $G$ is $15$ and from the orbit-stabilizer theorem, we know that the size of the orbit of an element $x \in X$ must divide the order of the group $G$. Therefore, the size of the orbit of an element $x \in X$ must be $1$, $3$, $5$, or $15$. Since the size of the set $X$ is $7$, the size of orbit can not be $15$.

Suppose that there are no fixed points of the action. This implies all orbits are of size $3$ or $5$. Since set of all orbits partitions the set $X$ and neither of $3$ or $5$ divide $7$, we can find $m,n \in \mathbb{N} $ such that \[ 3m + 5n = 7. \] But, the above does not have any solution in $\mathbb{N} \times \mathbb{N} $ as the minimum is $8$. Therefore, there must be some element of $X$ that is fixed by every element of $G$.