Every G-set can be thought of as a Category with an object(with morphisms elements of G), Category of sets (morphisms as set maps) and a functor between them. From this definition we get the G action which is a left action. So someone suggested that a contravariant functor from Category with on object and Category of sets will give the right action. Contravariant functor from Category of sets is same as a functor between them, since opposite of Category of sets is "itself".(Defn of contravariant functor from A to B is given as functor from A^op to B ). So contravariant functor will not give rise to right action. I think it is the the way we write functions that leads to only left actions for G-sets.
Loosely speaking, a right group action of G on a set S is a map S × G → S mapping (s, g) → s·g, satisfying (s·g)·h = s·(g·h).
Suppose we define F(g)·s := s·g and we know that F(g·h) = F(h)·F(g) because F is a contravariant functor.
Now see that s·(g·h) = F(g·h)·s = (F(h)·F(g))·s = F(h)·(F(g)·s) = F(h)(s·g) = (s·g)·h and therefore, right actions are contravariant functors.
Suppose we define F(g)·s := s·g and we know that F(g·h) = F(h)·F(g) because F is a contravariant functor.
Now see that s·(g·h) = F(g·h)·s = (F(h)·F(g))·s = F(h)·(F(g)·s) = F(h)(s·g) = (s·g)·h and therefore, right actions are contravariant functors.
In this regard, I want to quote a line from Tom Leinster's Basic Category Theory: "That left actions are covariant functors and right actions are contravariant functors is a consequence of a basic notational choice: we write the value of a function f at an element x as f(x), not (x)f."