Let \(G\) be a finite group and \(p\) a prime. Any two Sylow $p$-subgroups of \(G\) are conjugate: if \(P,Q\in\Syl_p(G)\), then there exists \(g\in G\) with \(gPg^{-1}=Q\). In particular, a Sylow \(p\)-subgroup \(P\) is normal in \(G\) if and only if it is the unique Sylow \(p\)-subgroup of \(G\) [@DF04].
Write \(\abs G=p^a m\) with \(p\nmid m\), and let \(P\) and \(Q\) be Sylow \(p\)-subgroups of \(G\). Let \(P\) act on the set \(G/Q\) of left cosets of \(Q\) by left translation, \(x\cdot gQ=xgQ\). Since \(\abs{G/Q}=m\) is prime to \(p\), the fixed-point congruence gives \(\abs{(G/Q)^P}\equiv m\not\equiv 0\pmod p\), so some coset \(gQ\) is fixed by \(P\). Then \(xgQ=gQ\) for every \(x\in P\), that is, \(g^{-1}Pg\subseteq Q\), and since both groups have order \(p^a\), \(g^{-1}Pg=Q\).
For the second statement, every conjugate \(gPg^{-1}\) has order \(p^a\) and is therefore a Sylow \(p\)-subgroup. If \(P\normal G\), every conjugate of \(P\) equals \(P\), so by the first statement \(P\) is the only Sylow \(p\)-subgroup. Conversely, if \(P\) is the only Sylow \(p\)-subgroup, then \(gPg^{-1}=P\) for every \(g\in G\).
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T-3X5FFSylow's third theorem: numerical constraintsuses