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Here is another solution:X is infinite and so there are two elemnts x1 and x2.Let X'=X{x1,x2}.I have that |X|=|X'|and so |P(X)|=|P(X')|.Let A be a subset of X' and so |X\A|>=2.Then there exist a permutation f:X\A----> X\A without fixed points.Then I extend f to X leaving fixed the elemnts of A.The set of the points fixed by f is then A.So I have a surjection Sym(X)---> P(X').Hence |Sym(X)|>=|P(X')|=|P(X)|.That |Sym(X)|<=|P(X)| is easy.