No. For any field $k$, there exists a pair of isogenous elliptic curves over $k$ that are not isomorphic. The dual isogeny yields suitable maps of function fields that are not isomorphisms. If your base field has finite transcendence degree and does not contain an algebraically closed subfield, then the same is true for the function fields.

Edit: I see that this answer is essentially identical to Robin Chapman's answer to a more general question, and to KConrad's answer to a slightly different question.

No. For any field $k$, there exists a pair of isogenous elliptic curves over $k$ that are not isomorphic. The dual isogeny yields suitable maps of function fields that are not isomorphisms. If your base field has finite transcendence degree and does not contain an algebraically closed subfield, then the same is true for the function fields.

Edit: I see that this answer is essentially identical to Robin Chapman's answer to a more general question, and to KConrad's answer to a slightly different question.

No. For any field $k$, there exists a pair of isogenous elliptic curves over $k$ that are not isomorphic. The dual isogeny yields suitable maps of function fields that are not isomorphisms. If your base field has finite transcendence degree and does not contain an algebraically closed subfield, then the same is true for the function fields.

No. For any field $k$, there exists a pair of isogenous elliptic curves over $k$ that are not isomorphic. The dual isogeny yields suitable maps of function fields that are not isomorphisms. If your base field has finite transcendence degree and does not contain an algebraically closed subfield, then the same is true for the function fields.

Edit: I see that this answer is essentially identical to Robin Chapman's answer to a more general question, and to KConrad's answer to a slightly different question.