= Solution
Compose <permutations> from right to left. The three <transpositions> are $\tau_1=(23)$, $\tau_2=(13)$ and $\tau_3=(12)$. In <cycle notation>, direct composition gives
$$
\tau_1\tau_2=(123),\qquad \tau_2\tau_1=(132),\qquad \tau_i^2=e.
$$
The complete <symmetric group> is therefore
$$
\boxed{S_3=\{e,\tau_1,\tau_2,\tau_3,\tau_1\tau_2,\tau_2\tau_1\}.}
$$
<Conjugation> relabels the entries of a <cycle>, so the <conjugacy classes> are precisely
$$
\boxed{\{e\},\qquad\{\tau_1,\tau_2,\tau_3\},\qquad\{\tau_1\tau_2,\tau_2\tau_1\}.}
$$
Their sizes are $1,3,2$, respectively. The differing <orders of group elements> also show that these three sets cannot merge under <conjugation>.
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