Can proper elementarily equivalent end extensions ever be definable?
There is a model of $PA$ that interprets a proper end extension with the same theory.
Let $M$ be an $\omega_1$-saturated model of true arithmetic. We will let elements of $n$ code (internally) finite sets in a typical way (such as $n \in m$ if and only if the $n$th binary digit of $m$ is $1$). Let $m$ be an element of $M$ such that for any standard natural $n$, $n \in m$ if and only if $M \models \varphi_n$, where $\varphi_n$ is the $n$th sentence in the language of arithmetic with some computable coding. Such an $m$ exists by $\omega_1$-saturation. Now consider the language $\mathcal{L}$ that is the language of arithmetic $\{0, 1,+,\cdot, <\}$ plus a single new constant $c$.
For any $k \in M$ less than or equal to the number of binary digits of $m$, let $T_k$ be the $\mathcal{L}$-theory that consists of $\varphi_n$ for any $n \in m$ together with the axioms $1+\dots+1 < c$ with $i$ many $1$'s for every positive $i \in M$. There is a single formula $\psi(k,n,m)$ such that $\varphi_n \in T_k$ if and only if $k \leq m$ and $M \models \psi(k,n,m)$.
Because $M$ is a model of true arithmetic, for any standard $k$, $M \models \mathrm{Con}(T_k)$. Therefore the largest $k$ less than or equal to the number of binary digits in $m$ such that $M \models \mathrm{Con}(T_k)$ is nonstandard. Let $k^\prime$ be the maximal such $k$.
Finally $PA$ (and therefore true arithmetic) is enough to prove that any arithmetically definable theory (with parameters) has an arithmetically definable model, specifically we can take the Henkinization of $T_{k^\prime}$ and then take the term model of the leftmost path of the tree of completions of that theory.
By construction this model is also a model of true arithmetic and is a proper end extension of $M$.
Moreover, with some basic model theory it is consistent that this interpreted structure is actually isomorphic to $M$, but obviously by Tarski's undefinability theorem, such an isomorphism can never be definable in the structure itself.