 Solution

For the nilpotent matrix $T$, define the polynomial $E(z)={\sum }_{k=0}^{m-1}{z}^k{T}^k/k!$. Because $T^m=0$, differentiating gives $E^{\prime }(z)=TE(z)=E(z)T$. The derivative of the preserved bilinear form matrix is thereforeIt is a constant matrix polynomial, equal to $A$ at $z=0$. Evaluation at one provesAlso $E(-1)$ is the inverse of $E(1)$ by the finite exponential product identity, so this really is an invertible isometry, even when $A$ is degenerate.