If $e$ is a small number, then $od(H)\approx od(A)+e(od)'_A(B)$. $od(A)=1-\dfrac{u(A)}{v(A)}$ where $u(A)=\det(A^*A),v(A)=\Pi_i||Ae_i||^2$ and $(e_i)_i$ is the canonical basis. $(od)'_A=-\dfrac{1}{v(A)}u'_A+\dfrac{u(A)}{v^2(A)}v'_A$. $u'_A(K)=trace((A^*K+K^*A)adjoint(A^*A))$. $v'_A(K)=\sum_i((e_i^*K^*Ae_i+e_i^*A^*Ke_i)\Pi_{j\not= i}||Ae_j||^2)$.
Edit:
- The proof is based on this fact: if $\phi:A\rightarrow \det(A)$, then $\phi'_A:K\rightarrow trace(K.adjoint(A))$.
- Simplifications: $u(A)=|\det(A)|^2$ and $e_i^*K^*Ae_i+e_i^*A^*Ke_i=2Real((K^*A)_{i,i})$.
- In the sake of simplicity, assume that $K$ is real. Let $(E_{i,j})$ be the canonical basis of $\mathcal{M}_n(\mathbb{R})$. Then the matrix of $(od)'_A$ is in the form $U=[u_1,\cdots,u_{n^2}]$. If you want a large variation of $od(A)$, then choose $B=U^T$ (in the orthogonal of $\ker(od'_A)$). Then $(od)'_A(B)=||U||^2$. For instance, if $A=\begin{pmatrix}8&-5-6I&-3-I\\-5+4I&2+3I&7+7I\\6-I&-4+6I&8+2I\end{pmatrix}$, then $U\approx [-0.0029,-0.0154,-0.0120,-0.0010,-0.0064,-0.0119,-0.0110,-0.0111,0.0124]$. $od(A)\approx 0.9489$ and $od(A-100.U^T)\approx 0.8147$.