Equivalence Class of Permutations Problem
I want to clarify some doubts that I'm having pertaining to the equivalence classes of permutations.
In my Abstract Algebra course, I'm being asked to find the equivalence class of $(1,2,3,4,5)$, which is an element of $S_5$, where $S_5$ is the set of all permutations of the set $\{1,2,3,4,5\}$. 
Is the equivalence class: $\{(1,2,3,4,5), (2,3,4,5,1), (3,4,5,1,2), (4,5,1,2,3), (5,1,2,3,4)\}$ , or are there more elements in this equivalence class?
 A: First of all: What does "View $S_3$ as a subset of $S_5$, in the obvious way." mean?
Every element $\sigma \in S_3$ is a bijection $\sigma \colon \{1,2,3\} \to \{1,2,3\}$. If we define $\sigma^* \colon \{1,2, \ldots, 5\} \to \{1,2, \ldots, 5\}$ by $\sigma^*(1) = \sigma(1), \sigma^*(2) = \sigma(2), \sigma^*(3) = \sigma(3), \sigma^*(4) = 4$ and $\sigma^*(5) = 5$, then $\sigma^* \in S_5$ and $\sigma^* \restriction \{1,2,3\} = \sigma$. In this way, we may identify any $\sigma \in S_3$ with $\sigma^* \in S_5$, which allows us to pretend that $S_3 \subseteq S_5$.
Now consider $\sigma = (1,2,3,4,5) \in S_5$. We say that $\tau \in S_5$ is equivalent to $\sigma$ iff $\sigma \circ \tau^{-1}(4) = 4$ and $\sigma \circ \tau^{-1}(5) = 5$. Since $\sigma(3) = 4$ and $\sigma(4) = 5$ this is equivalent to $\tau^{-1}(4) = 3$ and $\tau^{-1}(5) = 4$. But this is equivalent to $\tau(3) = 4$ and $\tau(4) = 5$.
Therefore the equivalence class of $\sigma = (1,2,3,4,5)$ contains precisely those $\tau \in S_5$ with $\tau(3) = 4$ and $\tau(4) = 5$. Let $\Sigma$ be the equivalence class of $\sigma$. Since $\tau \in \Sigma$ iff $\sigma \circ \tau^{-1} \in S_3$ iff $\tau^{-1} \in \sigma^{-1} \circ S_3$ and $\mid S_3 \mid = 6$, we know that $\mid \Sigma \mid = 6$.
Now $X = \{(3,4,5,1,2),(3,4,5,2,1),(3,4,5)(1,2),(3,4,5,1)(2),(3,4,5,2)(1),(3,4,5)(1)(2)\} \subseteq \Sigma$ and $\mid X \mid = 6$ (i.e. the elements listed in $X$ are pairwise distinct) and thus $X = \Sigma$.
