Without further requirements on $\tau$, this is trivial. Let $T$ be any first-order theory and let $\top$ be any tautology. Define $\tau(\phi)= \top$.

If you want $\tau$ to be injective, then let $T$ be any first-order theory in the language of $TA$ and define $\tau(\phi)=\top\vee\phi$.

If you also want equivalence, then take $T$ to be any first-order theory in the language of $TA$ (such that $T\subseteq TA$) and define $\tau$ as follows:

-$\tau(\phi)=\phi$ if $\phi$ is not a theorem of $TA$

-$\tau(\phi)=\top\vee\phi$ if $\phi$ is a theorem of $TA$

This $\tau$ is injective and gives you the equivalence:
$\forall \phi(TA\vdash\phi\leftrightarrow T\vdash\tau(\phi))$.

If you want $\tau$ to be an interpretation in the usual sense, as the title of the question suggests, then this is not possible as Fedor Pakhomov has explained in two comments (intepretations preserve negation).

So, you must specify what do you mean by a suitable $\tau$, as Monroe Eskew has already asked in a comment.

EDIT:

The edited version of the question suggests that the OP wants $\tau$ to be an interpretation in the usual sense. There is no such $\tau$ (this was already clear from the comments and the previous version of the answer, but I will unify the argument here for convenience).

Indeed, assume without loss of generality that the signature of $TA$ is finite, and let $\tau$ be an interpretation of $TA$ in $T$. In this case, $\tau$ is computable. Therefore, the set of sentences $\phi$ such that $T\vdash \tau(\phi)$ is RE. However, the set of those sentences is just the set of true sentences of arithmetic, for if $TA\vdash \phi$ then $T\vdash\tau(\phi)$, and if $TA\not\vdash\phi$, then $TA\vdash\neg\phi$, $T\vdash\neg\tau(\phi)$, and $T\not\vdash\tau(\phi)$. This is a contradiction with Tarski's theorem.

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