Functional Modulus

Clearly $f(x) = x$ is a solution.

Let $f(x) \neq x$ and let $f(x_{i}) = x_{j_{i}}$ ; which implies that $x_{i} \neq x_{j_{i}}$ (as it was our assumption that $f(x) \neq x$ ).

Now let $|f(x_{1}) - x_{1}| =|f(x_{2}) - x_{2}| = .......... =|f(x_{n}) - x_{n}| = k \neq 0$ .

So, $f(x_{1}) - x_{1} = \pm k$ , $f(x_{2}) - x_{2} = \pm k$ .......... $f(x_{n}) - x_{n} = \pm k$ .

Adding up all these equalities, $f(x_{1})+........+f(x_{n})-x_{1}.......-x_{n} = \pm k .....'n' times... \pm k$

But $f(x_{1}),f(x_{2}),.......f(x_{n})$ are permutations of $x_{1},x_{2},......x_{n}$ .

So L.H.S is equal to zero. But R.H.S cannot become zero because if $l$ k's are positive, then $n-l$ k's are negative. which forces that among $l$ and $n-l$ one is even and the other is odd.

So, there will be odd numbers of k's left over, which will not sum to zero.

So, only one one-one function exists.

$\therefore$ No. of functions = $\boxed{1}$ .