yes, you have to show that for any g in G, and h in H, that ghg^{-1} is in H.
equivalently (and easier to use for this problem), you need to show that gHg^{-1} is a subset of H, for any g in G.
some things to think about:
a) prove that for any subgroup H of G, and any element g in G, the set gHg^{-1} = {ghg^{-1}: h in H} is actually a subgroup of G.
b) show that H and gHg^{-1} have the same order.
conclude that if G has only ONE subgroup of a given order, that subgroup MUST be normal (sylow subgroup or not).
the definition of a sylow p-subgroup is, by the way:
a subgroup of order p^{k}, where p^{k} divides |G|, and p^{k+1} does not divide |G|.
for example, if G is a group of order 24, a sylow 2-subgroup of G is a subgroup of order 8, whereas a sylow 3-subgroup is a subgroup of order 3.