In this reference (Definition 2.47) a Banach space $X$ is called separably automorphic if given a separable space $Y$ and isomorphic copies $A$ and $B$ of $Y$ in $X$, every bijective operator $T:A\to B$ can be extended to an automorphism (bijective operator) $\hat T$ of $X$.

The spaces $\ell_\infty(\Gamma)$ are separably automorphic by Proposition 2.52 in the reference.

Let $Y$ be a separable subspace of $\ell_\infty(\Gamma)$. Taking a numerable subset $\Gamma_1$ of $\Gamma$, we can identify $\ell_\infty(\Gamma_1)$ with a copy of $\ell_\infty$ in $\ell_\infty(\Gamma)$. Moreover $\ell_\infty(\Gamma_1)$ contains an isometric copy $Y_1$ of $Y$. Given a bijective isometry $T:Y_1\to Y$ and the automorphic extension $\hat T$ we mentioned, $\hat T(\ell_\infty(\Gamma_1))\supset Y$.

This result gives a positive answer to the initial question.

The Lemma can be proved with the same arguments.

In this reference (Definition 2.47) a Banach space $X$ is called separably automorphic if given a separable space $Y$ and isomorphic copies $A$ and $B$ of $Y$ in $X$, every bijective operator $T:A\to B$ can be extended to an automorphism (bijective operator) $\hat T$ of $X$.

The spaces $\ell_\infty(\Gamma)$ are separably automorphic by Proposition 2.52 in the reference.

Let $Y$ be a separable subspace of $\ell_\infty(\Gamma)$. Taking a numerable subset $\Gamma_1$ of $\Gamma$, we can identify $\ell_\infty(\Gamma_1)$ with a copy of $\ell_\infty$ in $\ell_\infty(\Gamma)$. Moreover $\ell_\infty(\Gamma_1)$ contains an isometric copy $Y_1$ of $Y$. Given a bijective isometry $T:Y_1\to Y$ and the automorphic extension $\hat T$ we mentioned, $\hat T(\ell_\infty(\Gamma_1))\supset Y$.

This result gives a positive answer to the initial question.

The Lemma can be proved with the same arguments.

ADDED on June 28, 2024: The above arguments prove the isomorphic version of the problem, but not the isometric version: in general $T$ isometry does not imply $\hat T$ isometry.