Will the two elements in $S_n$ have the same cycle structure?

Question

Let $G$ be a finite group. Let $a,b\in G$ such that $o(a)=o(b)$.

Since any group $G$ is isomorphic to a subgroup of $S_n$ for some $n$ then there exists a monomorphism from $G$ to $S_n$ .

Let the elements be mapped to $a_1,b_1$.

Question:Will $a_1,b_1$ have the same cycle structure?

Now $a_1,b_1$ can be expressed as a product of disjoint cycles $a_1=\sigma_1\sigma_2\dots \sigma_n;b_1=\rho_1\rho_2\dots \rho_m$. Also $o(a_1)=o(b_1)$.

EDIT: The original question was :

If $a,b\in G$ be such that $o(a)=o(b)$ then does there exist a group $H$ containing $G$ such that $a,b$ are conjugate in $H$.

I took $H=S_n$ and we know that two elements are conjugate in $S_n\iff$ they have the same cycle structure.

Is the original question false then?

Answer 1

Taking $G=S_n$ itself, your question boils down to asking whether two elements in $S_n$ with the same order have the same cycle structure. This is, of course, not true: Take $$ n=4, \sigma = (12), \tau = (12)(34) $$