Infinite descending chain of Turing jumps with equality

Let $\Theta(X, n)$ be a uniform process such that $\Theta(Z^{(n)}, n) = Z$, i.e. $\Theta$ undoes jumps. Using the recursion theorem, define a functional $\Phi_e$ such that $\Phi_e^X(e)\downarrow$ iff $\exists n\, \Theta(X, n)(e) = 0$.

Now suppose such a sequence $(X_i)_{i \in \omega}$ existed. Consider the sequence $X_0(e), X_1(e), X_2(e),\dots$. So $X_i(e) = 1$ iff there is a $j \ge i$ with $X_j(e) = 0$. If this sequence contains infinitely many 0s, then it is is entirely 1s, a contradiction. If this sequence contains only finitely many 0s, then it contains only finitely many 1s, again a contradiction.

This is a consequence of the main result of John Steel's paper Descending Sequences of Degrees. The main result of that paper is that if $P$ is an arithmetic formula, there is no sequence $\langle A_n \rangle_n$ such that for all $n$, $A_n \ge_T A_{n + 1}'$ and $A_{n + 1}$ is the unique $B$ such that $P(A_n, B)$ holds. In your case, you just need to let $P(X, Y)$ be the formula which says that $X$ is the jump of $Y$ (since the jump, at least in its usual formulation, is injective on the level of sets).

If you ask instead about Turing equivalence, then your statement is false. Here's a proof. Let $\langle A_n \rangle_n$ be a sequence such that $A_n \ge_T A_{n + 1}'$. Then you can use the relativized form of the jump inversion theorem to build a sequence $\langle B_n \rangle_n$ such that for each $n$, $B_n \ge_T A_n$ and $B_n \equiv_T B_{n + 1}'$.