If ψ is an automorphism and we know ψ(X) then ψ(x) is determined for all other x.

As you can see this notation is very confusing. I shall attempt to see if lambda notation works. I recommence:

There is a short proof that in GF(p^{q}) φ = λx.x^{p} is an automorphism.
Within the group of automorphisms, φ has order q for φ^{q} is the identity; i.e. x^{pq} = x for all x.

If ψ is an automorphism then ψ(λx.x) entirely determines ψ. Given ψ(λx.x), ψ(y) = ξ