This question originates from Pinter's Book of Abstract Algebra Chapter 16 exercise N3.
Let $G$ be a finite group, and $K$ a p-Sylow subgroup of $G$. Let $X$ be the set of all the conjugates of $K$. If $C_1, C_2\in X$, let $C_1\sim C_2$ iff $C_1 = aC_2a^{-1}$ for some $a\in K$. Thus, $\sim$ partitions $X$ into equivalence classes. If $C\in X$, let the equivalence class of $C$ be denoted by $[C]$. Proves the only class with a single element is $[K]$.
Here is a proof:
Every conjugate of a p-Sylow subgroup of $G$ is a p-Sylow subgroup of $G$ (by Chapter 16 exercise M2). Thus every element $C\in X$ is a p-Sylow subgroup of $G$.
By definition, $[C] = \{aCa^{-1}: a\in K\}$ and $N_K(C) = \{a\in K: aCa^{-1} = C\}$. Note $N_K(C)$ is a subgroup of $K$ (by Chapter 14 exercise I5).
Let $C^* = \{N_K(C)a: a\in K\}$. By Chapter 14 exercise I10, $[C]$ is in one-to-one correspondence with $C^*$, and the number of elements in $[C]$ is a divisor of $K$. In particular, $|[C]| = |C^*| = (K: N_K(C))$, by Lagrange's Theorem.
$a\in K$ implies $a$ has order a power of $p$ (for $K$ is a p-Sylow subgroup of $G$), and $aCa^{-1} = C\implies a\in N_K(C)$. It follows $a\in C$, by Chapter 16 exercise M7. Hence, $N_K(C)\subseteq (K\cap C)$.
As $C$ is a conjugate of $K$, $|C| = |K|$ (by Chapter 14 exercise I2). It's easy to see that $|[K]| = 1$ (as $aKa^{-1} = K$ for all $a\in K$). Suppose $C\ne K$. Then $(K\cap C)\subset K$, and therefore $|N_K(C)| < |K|$. It follows that $|[C]| = (K: N_K(C)) > 1$.
Does it look right? Is there a simpler proof?