The number of the sets, $A$ s.t. $f^{-1}(f(A))=A$

Say $f colon X to Y$ such that $1mapsto2, 2mapsto3, 3mapsto4, 4mapsto3, 5mapsto1, 6mapsto2, 7mapsto3$.

(Here the $X ={1,2,3,4,5,6,7}, Y = {1,2,3,4,5,6}$)

Find the number of the sets $A$ satisfying $f^{-1}(f(A))=A$

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In my thought, To claim $f^{-1}(f(A))=A$, $f$ should be one to one function(or injective) for the set $A$.

So considering $f({1,6}) = {2}$, $f({2,4,7}) = {3}$, $f({3}) = {4}$ and $f({5}) = {1}$

So I classified the case like the below for making the injective for $f : A to Y$

$(1)$ $f({5}) = {1}$ ; There are $2$ cases $5$ is element of the $A$ or not.

$(2)$ $f({3}) = {4}$ ; There are $2$ cases $4$ is element of the $A$ or not.

$(3)$ $f({2,4,7}) = {3}$; There are $4$ case, because $A$ should be include only one element among ${2,4,7}$ or not include any element.

$(4) f({1,6}) = {2}$ ; with the same logic $A$ can be $3$ cases.

So my answer is $ 2times 2 times 4 times 3 =48$ But the answer was 16. What the point did I have mistake? Thanks.