FALSE: an algebraic closure is unique up to a unique base-field isomorphism

Choose $\mathrm{\Omega }$ by [L1]. The polynomial $x^2+1$ is irreducible over $\mathbb{Q}$ and has a root $a\in \mathrm{\Omega }$ by [L5]. Its roots $a$ and $-a$ are distinct because $\mathbb{Q}$ has characteristic zero. By [L2], the assignment $a\mapsto -a$ gives a nonidentity $\mathbb{Q}$-embedding of $\mathbb{Q}(a)$ into $\mathrm{\Omega }$.

Extend that embedding across the algebraic extension $\mathrm{\Omega }\mathrm{/}\mathbb{Q}(a)$ using [L3]. Its image $E$ is algebraically closed because it is isomorphic to $\mathrm{\Omega }$. Every $b\in \mathrm{\Omega }$ is algebraic over $\mathbb{Q}\subseteq E$ by [L5], so [L6] gives a minimal polynomial over $E$; it has a root in $E$ and is therefore linear. Thus the resulting embedding $\tau :\mathrm{\Omega }\to \mathrm{\Omega }$ is surjective, hence is a nonidentity $\mathbb{Q}$-automorphism with $\tau (a)=-a$.

The identity and $\tau$ are distinct $\mathbb{Q}$-isomorphisms from the same algebraic closure to itself. Therefore uniqueness of the base-field isomorphism is false, although existence is true by [L4].