Proof:

Claim: H = {x ∈ G | ϕ(x) = x} is a subgroup of G.

To prove that H is a subgroup of G, we need to show that it satisfies three conditions: closure, identity, and inverses.

Closure:
Let a, b ∈ H. We need to show that ab^(-1) ∈ H.

Since a, b ∈ H, we have ϕ(a) = a and ϕ(b) = b.

Since ϕ is an automorphism, it preserves the group operation, so ϕ(a)ϕ(b^(-1)) = ϕ(ab^(-1)).

Therefore, ab^(-1) ∈ H, and H is closed under the group operation.

Identity:
Since ϕ is an automorphism, it preserves the identity element. Let e be the identity element of G.

We have ϕ(e) = e, so e ∈ H.

Inverses:
Let a ∈ H. We need to show that a^(-1) ∈ H.

Since a ∈ H, we have ϕ(a) = a.

Since ϕ is an automorphism, it is bijective and has an inverse ϕ^(-1).

Applying ϕ^(-1) to both sides of the equation ϕ(a) = a, we get ϕ^(-1)(ϕ(a)) = ϕ^(-1)(a), which simplifies to a = ϕ^(-1)(a).

Therefore, a^(-1) = ϕ^(-1)(a) ∈ H, and H contains inverses.

Conclusion:
Since H satisfies closure, identity, and inverses, it is a subgroup of G.

Summary:
To prove that H = {x ∈ G | ϕ(x) = x} is a subgroup of G, we showed that it satisfies the three conditions for being a subgroup: closure, identity, and inverses. This was done by utilizing the properties of automorphisms, specifically that they preserve the group operation, the identity element, and have an inverse. Therefore, H is indeed a subgroup of G.