Pfaffian equals complex determinant?

Let $V$ be a Euclidean vector space and let $V^{\mathbb{C}} = V \oplus V$ be its complexification, with complex structure $$J = \begin{pmatrix} 0 & -\mathrm{id}\\ \mathrm{id} & 0 \end{pmatrix}.$$ Of course, we can regard $V^{\mathbb{C}}$ also as a real vector space with a canonical orientation (for a basis $v_1, \dots, v_n$ of $V$, the basis $(v_1, 0), (0, v_1), \dots, (v_n, 0), (0, v_n)$ is positively oriented). Now let $A$ be an anti-symmetric endomorphism of $V$ and consider the anti-symmetric endomorphism $$\tilde{A} = \begin{pmatrix} A& -\mathrm{id}\\ \mathrm{id} & A\end{pmatrix}$$ of $V^{\mathbb{C}}$. Then $\tilde{A}$ has a well-defined Pfaffian. On the other hand, we have $\tilde{A} = A + i$. Is it true that $$\mathrm{Pf}(\tilde A) = \det(A+i),$$ at least if $V$ is even-dimensional? What about the odd-dimensional case?

The answer is YES in every dimension, up to a sign. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore $${\rm Pf}(\tilde A)^2=\det(I+iA)\det(I-iA).$$ On the other hand $$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^*=\det(I+iA),$$ yields $${\rm Pf}(\tilde A)^2=(\det(I+iA))^2.$$ Let us remark that $\det(I+iA)$ is a polynomial in the entries of $A$, with real entries because $I+iA$ is Hermitian. We deduce $${\rm Pf}(\tilde A)=\epsilon\det(I+iA)$$ for some constant $\epsilon=\pm1$. Taking $A=0_n$ gives $\epsilon=1$.
Remark that ${\rm Pf}(\tilde A)$, as a polynomial in the entries of $A$, is even. In particular, if $n$ is odd, ${\rm Pf}(\tilde A)$ has degree $n-1$ only, whereas if $n$ is even, then ${\rm Pf}(\tilde A)$ has degree $n$. For instance, if $n=3$ and the off-diagonal entries of $A$ are $\pm a,\pm b,\pm c$, then $${\rm Pf}(\tilde A)=1-a^2-b^2-c^2.$$