Given an isomorphism $e : U \oplus U' \cong V$ and automorphisms $\alpha$ of $U$, $\beta$ of $U'$, the conjugated map $h = e \circ (\alpha \oplus \beta) \circ e^{-1}$ has determinant $\det h = \det \alpha \cdot \det \beta$.
Specialization: if $\det \beta = (\det \alpha)^{-1}$, then the extended automorphism $h$ has determinant $1$, lying in the special isometry group.
The extended automorphism $h$ restricted to the Lagrangian subspace $e(U \times \{0\})$ acts as $\alpha$: $h(e(u, 0)) = e(\alpha u, 0)$.
Lagrangian extension is an isometry: given that $e(U \times \{0\})$ and $e(\{0\} \times U')$ are isotropic Lagrangians and $\alpha, \beta$ are mutually adjoint, the extended map $h = e \circ (\alpha \oplus \beta) \circ e^{-1}$ preserves the bilinear form $B$.