Does this exist? I don't think it does. For any cyclic group the totient function of 10 is 4, so there is 4 of them. But also if one element is of order 10, say $a$, then $a^3$, $a^7$ is also of order 10 but thats three of them. Now I understand what I just did is mainly for cyclic groups, but can I apply that idea to non cyclic groups such as dihedrals or symmetric groups?
can I apply that idea to non cyclic groups such as dihedrals or symmetric groups?
Yes. If you have an element of order 10 in any group, it generates a subgroup of order 10, which is cyclic by definition. Since this subgroup has four generators, each of those is an element of order 10 in the subgroup, and is also of order 10 in the original group.
If $x$ is an element of order $10$, then also $x^3, x^7, x^9$ are all different elements of order 10. So it is impossible to have exactly two of them.
In general, if $n$ is a natural number and $G$ is a group with exactly $2$ elements of order $n$, then $n=3, 4$ or $6$. This follows from solving the equation $\varphi(n)=2$, where $\varphi$ is Euler's totient function. Each of the cases can be realized: take $G \cong D_n$, with $n=3,4$ or $6$.
Note that $D_3\cong S_3$ and $D_6\cong S_3 \times C_2$.