$
\newcommand\ZFC{\text{ZFC}}
\newcommand\HOD{\text{HOD}}
\newcommand\Ord{\text{Ord}}
$

I like this question. Here are some set-theoretic counterexamples,
which may be closer to what you are thinking about than Bjorn's
nice example.

**Theorem.** Assume ZFC + V=HOD plus $V\neq L$ plus $0^\sharp$
does not exist; that is, in $\langle V,\in\rangle$. Then there is
a definable relation $\in'$ on $V$ such that $\langle
V,\in'\rangle$ has no nontrivial elementary embeddings, yet $\in$
is not definable from $\in'$.

Proof. Since V=HOD, it follows that there is a definable bijection
of $V$ with $\Ord$, and also of $\Ord$ with $L$. Thus, we get a
definable bijection $\pi:V\to L$. Define $x\in' y\leftrightarrow \pi(x)\in
\pi(y)$, so that $\pi$ is a definable isomorphism of $\langle
V,\in'\rangle$ with $\langle L,\in\rangle$. Since $0^\sharp$ does
not exist, it follows that there are no nontrivial
$\in'$-elementary embeddings $j:V\to V$, since there are no nontrivial $\in$-embeddings $j:L\to L$. But meanwhile, $\in$ is
not definable from $\in'$ in $V$, since if it were, we could run
the definition inside $\langle L,\in\rangle$ to recover a copy of
$V$, which is impossible as this would give us access in $L$ to
sets in $V$ that are not in $L$. QED

The argument can be generalized to other inner models $K$ rather
than $L$, using the appropriate sharps, and indeed, we can avoid
the sharp issue entirely with the following:

**Theorem.** Every model $V$ of ZFC has a class forcing extension
$V[G]$ in which there is a definable relation $\in'$, such that
$\langle V[G],\in'\rangle$ admits no nontrivial self-embeddings and $\in^{V[G]}$ is not definable from $\in'$ from
parameters.

Proof. Let $V[G]$ be a nontrivial extension of $V$ satisfying
$V=\HOD$, obtained by the usual progressively closed iteration. It
follows that $V$ is a definable class in $V[G]$, and we may
actually arrange that this definition does not require parameters.
Thus, the extension has a definable bijection $\pi:V[G]\to V$, and
we may define $x\in' y\iff \pi(x)\in \pi(y)$. Thus, $\pi$ is an
isomorphism of $\langle V[G],\in'\rangle$ with $\langle
V,\in\rangle$. This forcing extension $V[G]$ can have no nontrivial elementary embedding $\langle V,\in\rangle\to\langle V,\in\rangle$, and hence no nontrivial elementary self embedding of $\langle V[G],\in'\rangle$. Yet, $\in^{V[G]}$ is not definable from $\in'$, even
with parameters, since if it were, we could run the definition
inside $\langle V,\in\rangle$, with $\pi$ of the parameters, and thereby get access to the sets
of $V[G]$ from $V$ itself, which is impossible. QED