Official solution
f(1) = 1 - 5\[\left[ {{1 \over 5}} \right]\] = 1
f(6) = 6 - 5\[\left[ {{6 \over 5}} \right]\] = 1
So, this function is many to one.
f(10) = 10 - 5\[\left[ {{10 \over 5}} \right]\] = 0 which is not present in the set of natural numbers.
So this function is neither one-one nor onto.