Diagonal lemma from recursion theorem?

Question

Does Gödel's diagonal lemma follow from Kleene's recursion theorem? I believe the converse is true, by an argument like the following.

Let e ↦ θ_e be a bijection between ω and the set of Σ⁰₁ formulas in the language of arithmetic having just the variable v₀ free.

Let e ↦ W_e be a numbering of the r.e. subsets of ω such that for all e, W_e = {n ∈ ω : θ_e(n) is true}. This is possible because every r.e. set is defined in the standard model of arithmetic by a Σ⁰₁ formula.

Let f be any recursive operation on ω. To prove the recursion theorem, we seek a number d such that W_d = W_f(d).

Toward that, let φ(v₀, v₁) be a Σ⁰₁ formula satisfied by a pair (n, e) iff θ_f(e)(n) is true. For instance, φ might be

Σ⁰₁-Tr[ Sub(numeral(v₀), 0, f(v₁)) ]

so that φ(n, e) says that a true Σ⁰₁ sentence results from substituting the numeral for n into the formula with code f(e).

By the diagonal lemma, there is a Σ⁰₁ formula θ_d(v₀) such that

∀v₀[ θ_d(v₀) ↔ φ(v₀, d) ]

is provable in Robinson's arithmetic. Hence

W_d = {n ∈ ω : θ_d(n) is true} = {n ∈ ω : φ(n,d) is true} = {n ∈ ω : θ_f(d)(n) is true} = W_f(d)

as desired. The second equality follows by choice of θ_d.

My question is whether we can prove the diagonal lemma from the recursion theorem. For specificity, here is the lemma I have in mind:

For any formula φ(v₀, v₁, … v_n) in the language of arithmetic, one can find a formula ψ(v₁, … v_n) in that language, so that Robinson's arithmetic proves ∀φv₁…∀φv_n[ ψ(v₁, … v_n) ↔ φ(⌜ψ⌝, v₁, … v_n) ].

Note that it is not sufficient to prove that for each φ there is a ψ for which the stated equivalence holds; the equivalence must be provable in Robinson's arithmetic.