Proof by Contradiction:
Let's assume that the mapping f: G → G such that f(x) = x^(-1) is an automorphism.

Definition of an Automorphism:
An automorphism is a bijective mapping from a set onto itself that preserves the structure of the set. In this case, the set is the group G.

Properties of an Automorphism:
1. f is a bijection: It means that f is both injective (one-to-one) and surjective (onto).
2. f preserves the group operation: It means that for any elements a and b in G, f(a * b) = f(a) * f(b).

Proof:
Since f is an automorphism, it must be a bijection. Therefore, every element in G must have a unique inverse under f.

Counterexample: Non-Abelian Group
Consider a non-Abelian group G. By definition, G is not commutative, which means that there exist elements a and b in G such that a * b is not equal to b * a.

Now, let's consider the mapping f(x) = x^(-1) for this non-Abelian group G. Since G is non-Abelian, there exist elements a and b such that a * b is not equal to b * a.

Proof of Non-Automorphism:
1. f is a bijection: Since every element in G has a unique inverse, f is injective (one-to-one). Also, since f maps each element to its inverse, it covers all elements in G, making it surjective (onto). Therefore, f is a bijection.
2. f preserves the group operation: Let's consider the product of two elements a and b in G: a * b. According to the group operation, the product of a and b is not equal to the product of b and a, i.e., a * b ≠ b * a.

Now, let's apply the mapping f to both sides of this equation:
f(a * b) = f(a) * f(b)
f(a * b) = a^(-1) * b^(-1)

Since a * b ≠ b * a, f(a * b) = a^(-1) * b^(-1) ≠ b^(-1) * a^(-1).

This contradicts the property that f preserves the group operation, and therefore, the mapping f: G → G such that f(x) = x^(-1) is not an automorphism for a non-Abelian group G.