Here is a counterexample which, unlike JSpecter's, does not rely on transcendence degrees and Zorn's lemma (used to create an embedding of ${\mathbf C}(X)$ into $\mathbf C$). It comes from a pair of isogeneous but nonisomorphic elliptic curves. Let $E_1$ and $E_2$ be elliptic curves over ${\mathbf Q}$ admitting an isogeny $E_1 \rightarrow E_2$. There is a (dual) isogeny $E_2 \rightarrow E_1$ and these maps provide one with homomorphisms between the function fields ${\mathbf Q}(E_2) \rightarrow {\mathbf Q}(E_1)$ and ${\mathbf Q}(E_1) \rightarrow {\mathbf Q}(E_2)$. So each function field embeds into the other. Choosing $E_1$ and $E_2$ to be isogeneous but not isomorphic over ${\mathbf Q}$ -- say their $j$-invariants are not equal to assure non-isomorphism -- the function fields ${\mathbf Q}(E_1)$ and ${\mathbf Q}(E_2)$ are not isomorphic as fields. This is an absolute statement since we work over ${\mathbf Q}$ rather than, say, ${\mathbf C}$ (where one would conclude the function fields are not isomorphic over ${\mathbf C}$, i.e., fixing ${\mathbf C}$). The same argument goes through using isogenous but non-isomorphic elliptic curves over ${\mathbf F}_p$ for any prime $p$.

By the way, this question is similar to Schroeder-Bernstein for Rings and my answer is similar to one given there.

Here is a counterexample which, unlike JSpecter's, does not rely on transcendence degrees and Zorn's lemma (used to create an embedding of ${\mathbf C}(X)$ into $\mathbf C$). It comes from a pair of isogeneous but nonisomorphic elliptic curves. Let $E_1$ and $E_2$ be elliptic curves over ${\mathbf Q}$ admitting an isogeny $E_1 \rightarrow E_2$. There is a (dual) isogeny $E_2 \rightarrow E_1$ and these maps provide one with homomorphisms between the function fields ${\mathbf Q}(E_2) \rightarrow {\mathbf Q}(E_1)$ and ${\mathbf Q}(E_1) \rightarrow {\mathbf Q}(E_2)$. So each function field embeds into the other. Choosing $E_1$ and $E_2$ to be isogeneous but not isomorphic over ${\mathbf Q}$ -- say their $j$-invariants are not equal to assure non-isomorphism -- the function fields ${\mathbf Q}(E_1)$ and ${\mathbf Q}(E_2)$ are not isomorphic as fields. This is an absolute statement since we work over ${\mathbf Q}$ rather than, say, ${\mathbf C}$ (where one would conclude the function fields are not isomorphic over ${\mathbf C}$, i.e., fixing ${\mathbf C}$). The same argument goes through using isogenous but non-isomorphic elliptic curves over ${\mathbf F}_p$ for any prime $p$.

By the way, this question is similar to Schroeder-Bernstein for Rings and my answer is similar to one given there.

Here is a counterexample which, unlike JSpecter's, does not rely on transcendence degrees and Zorn's lemma (used to create an embedding of ${\mathbf C}(X)$ into $\mathbf C$). It comes from a pair of isogeneous but nonisomorphic elliptic curves. Let $E_1$ and $E_2$ be elliptic curves over ${\mathbf Q}$ admitting an isogeny $E_1 \rightarrow E_2$. There is a (dual) isogeny $E_2 \rightarrow E_1$ and these maps provide one with homomorphisms between the function fields ${\mathbf Q}(E_2) \rightarrow {\mathbf Q}(E_1)$ and ${\mathbf Q}(E_1) \rightarrow {\mathbf Q}(E_2)$. So each function field embeds into the other. Choosing $E_1$ and $E_2$ to be isogeneous but not isomorphic over ${\mathbf Q}$, the function fields ${\mathbf Q}(E_1)$ and ${\mathbf Q}(E_2)$ are not isomorphic as fields. This is an absolute statement since we work over ${\mathbf Q}$ rather than, say, ${\mathbf C}$ (where one would conclude the function fields are not isomorphic over ${\mathbf C}$, i.e., fixing ${\mathbf C}$). The same argument goes through using isogenous but non-isomorphic elliptic curves over ${\mathbf F}_p$ for any prime $p$.

Here is a counterexample which, unlike JSpecter's, does not rely on transcendence degrees and Zorn's lemma (used to create an embedding of ${\mathbf C}(X)$ into $\mathbf C$). It comes from a pair of isogeneous but nonisomorphic elliptic curves. Let $E_1$ and $E_2$ be elliptic curves over ${\mathbf Q}$ admitting an isogeny $E_1 \rightarrow E_2$. There is a (dual) isogeny $E_2 \rightarrow E_1$ and these maps provide one with homomorphisms between the function fields ${\mathbf Q}(E_2) \rightarrow {\mathbf Q}(E_1)$ and ${\mathbf Q}(E_1) \rightarrow {\mathbf Q}(E_2)$. So each function field embeds into the other. Choosing $E_1$ and $E_2$ to be isogeneous but not isomorphic over ${\mathbf Q}$ -- say their $j$-invariants are not equal to assure non-isomorphism -- the function fields ${\mathbf Q}(E_1)$ and ${\mathbf Q}(E_2)$ are not isomorphic as fields. This is an absolute statement since we work over ${\mathbf Q}$ rather than, say, ${\mathbf C}$ (where one would conclude the function fields are not isomorphic over ${\mathbf C}$, i.e., fixing ${\mathbf C}$). The same argument goes through using isogenous but non-isomorphic elliptic curves over ${\mathbf F}_p$ for any prime $p$.

By the way, this question is similar to Schroeder-Bernstein for Rings and my answer is similar to one given there.