Suppose that $A \leq_a 0^\omega$ (i.e. $A$ is arithmetic in $0^\omega$) does there exist $\widehat{A} \equiv_a A$ with $\widehat{A} \leq_T 0^\omega$ [1]?

More generally, say that a set $X$ is aT-complete (better name?) if every arithmetic degree $\mathbf{a} \leq_a X$ has a representative $A \leq_T X$.  Does every arithmetic degree contain an aT-complete set?  Are there any non-trivial degrees containing aT-complete sets (clearly any arithmetic degree bounding only finitely many arithmetic degrees will contain an aT-complete set [2])

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[1]: In other words, if $A \leq_T 0^{\omega + n}$ does there exist $\widehat{A} \leq_T 0^\omega$, $m \in \omega$ such that $\widehat{A}^m \geq_T A$ and $A^m \geq_T \widehat{A}$?

[2]:  Suppose the only arithmetic degrees (strictly) below $\mathbf{a}$ are $b_1, \ldots, b_n$ with representatives $A, B_1, \ldots, B_n$ respectively.  Let $m_i$ be such that $A^{m_i} \geq_T B_i$ and $m = \max m_i$.  Then $A^{m}$ is aT-complete.