My attempt is below. Could I please get feedback on it. I am not so sure that it is correct.
Let α,β,σ ∈$S_n$.
Since $S_n$ is a group, we know that it contains an identity.
Let e be the identity. So, $\sim$ is reflexive since $\alpha = e \alpha e^{-1} = e \alpha = \alpha e^{-1} = \alpha$.
Since σ ∈ $S_n$ we know that there is a σ such that $σασ^{−1} = β$.
Further, because $S_n$ is a group, it must contain an inverse.
So, $α = σ^{−1}σασ^{−1}σ = σ^{−1}βσ$.
We now have $α = σ^{−1}βσ$ and $σασ^{−1} = β$, proving symmetry.
For transitivity, we must show that if α ∼ β and β ∼ c, then α ∼ c.
We already know that α ∼ β from above.
We still need to show that $σβσ^{−1} = c$.
Substituting in for β, $σ_1(σ_2ασ_2^{-1})σ_1^{-1} = (σ_1σ_2)α(σ_2^{-1} σ_1^{-1})$.
So, $σ_1^{-1} σ_2^{-1} = |σ_2σ_1|^{-1}$ as we are required to have a σ that shows $σβσ^{-1} = c$.
Reflexivity: Observe that for all $\alpha\in S_n$: $\alpha=e\cdot\alpha\cdot e^{-1}$, for $e$ the identity element of $S_n$. Thus $\alpha\sim\alpha$.
Symmetry: Suppose $\alpha\sim\beta$, then there is $\sigma\in S_n$ such that $\alpha=\sigma\beta\sigma^{-1}$, so we can conclude that $\beta=\sigma^{-1}\alpha\sigma=\sigma^{-1}\alpha(\sigma^{-1})^{-1}$, so $\beta\sim\alpha$.
Transitivity: Suppose $\alpha\sim\beta$ and $\beta\sim\gamma$. Then there exists $\sigma,\tau\in S_n$ such that $$ \begin{cases} \alpha=\sigma\beta\sigma^{-1} &(1)\\ \beta=\tau\gamma\tau^{-1} &(2) \end{cases} $$ so by replacing $\beta$ from equation $(2)$ in equation $(1)$ we will get $\alpha=\sigma\tau\gamma\tau^{-1}\sigma^{-1}=\sigma\tau\gamma(\sigma\tau)^{-1}$.