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Hey Damien, I think the following should work $\mathbb{C}$ and $\mathbb{C}(x)$. There is only one uncountable algebraically closed field of each cardinality in characteristic 0 and the algebraic closure of the right hand guy should have cardinality the continuum it should be isomorphic to $\mathbb{C}$. Probably, this assumes the axiom of choice though.

1

Hey Damien, I think the following should work $\mathbb{C}$ and $\mathbb{C}(x)$. There is only one algebraically closed field of each cardinality in characteristic 0 and the algebraic closure of the right hand guy should have cardinality the continuum it should be isomorphic to $\mathbb{C}$. Probably, this assumes the axiom of choice though.