For any such $K$, we can recover the subfield of constants $\mathbb{C}\subset K$ as the elements of $K$ that have $n$th roots for all $n$. Indeed, if a rational function on a curve has roots of all orders, it must have valuation $0$ at every point and hence be constant. Thus any isomorphism between two such fields $K$ and $K'$ must fix $\mathbb{C}$ setwise, and so can be described as an automorphism of $\mathbb{C}$ followed by an isomorphism of curves. That is, two such fields are isomorphic iff the corresponding curves are conjugate under some automorphism of $\mathbb{C}$. In particular, for example, this means that the genus is invariant under all such isomorphisms.

For any such $K$, we can recover the subfield of constants $\mathbb{C}\subset K$ as the elements of $K$ that have $n$th roots for all $n$. Indeed, if a rational function on a curve has roots of all orders, it must have valuation $0$ at every point and hence be constant. Thus any isomorphism between two such fields $K$ and $K'$ must fix $\mathbb{C}$ setwise, and so can be described as an automorphism of $\mathbb{C}$ followed by an isomorphism of curves. That is, two such fields are isomorphic iff the corresponding curves are conjugate under some automorphism of $\mathbb{C}$. In particular, for example, this means that the genus is invariant under all such isomorphisms (as any of the usual algebro-geometric definitions of genus are preserved by automorphisms of the base field).

For any such $K$, we can recover the subfield of constants $\mathbb{C}\subset K$ as the elements of $K$ that have $n$th roots for all $n$. Indeed, if a meromorphic function on a compact Riemann surface has roots of all orders, it must have valuation $0$ at every point and hence be constant. Thus if $K$ and $K'$ are isomorphic as abstract fields, they are also isomorphic over $\mathbb{C}$ (which happens iff the corresponding Riemann surfaces are isomorphic).

For any such $K$, we can recover the subfield of constants $\mathbb{C}\subset K$ as the elements of $K$ that have $n$th roots for all $n$. Indeed, if a rational function on a curve has roots of all orders, it must have valuation $0$ at every point and hence be constant. Thus any isomorphism between two such fields $K$ and $K'$ must fix $\mathbb{C}$ setwise, and so can be described as an automorphism of $\mathbb{C}$ followed by an isomorphism of curves. That is, two such fields are isomorphic iff the corresponding curves are conjugate under some automorphism of $\mathbb{C}$. In particular, for example, this means that the genus is invariant under all such isomorphisms.