We can check injectivity and surjectivity locally, hence we are reduced to the following algebraic situation:
$(R,\mathfrak m)$ is a noetherian local ring and $F,E$ two finitely generated $R$-modules with $E$ free. The map $f: F \to E$ is an isomorphism after tensoring with $R/\mathfrak m$, hence it is surjective by Nakayama.
But now, note that $E$ is free, in particular projective, hence the map splits, say by $g: E \to F$. After tensoring with $R/\mathfrak m$, $g \otimes R/\mathfrak m$ is an isomorphism, since it is the splitting of the isomorphism $f \otimes R/\mathfrak m$.
Again, by Nakayama, we deduce that $g$ is surjective. $g$ was a priori injective, hence $g$ is an isomorphism, thus $f$ is an isomorphism, too.