Let $(A,m)$ be a commutative local noetherian ring, and suppose that $(x_1,...,x_\ell)$ is a regular sequence inside the maximal ideal $m$. To this regular sequence, you can associate a canonical finite resolution of $A/(x_1,...,x_\ell)$ by free $A$-modules, called the Koszul resolution: $$ K(\mathbf{x}):0 \to \bigwedge^\ell A^\ell \overset{d_\ell} \to \bigwedge^{\ell-1} A^\ell \to \cdots \to \bigwedge^1 A^\ell \overset{d_1}\to A \to 0$$ where the differential is $$d_k (e_1 \wedge \dots \wedge e_k) = \sum_{i=1}^k (-1)^{i+1} d_1(e_i) e_1 \wedge \cdots \wedge \widehat{e_i} \wedge \cdots \wedge e_k.$$ The Koszul resolution also admits a sort-of "self-duality" (see here). That is, if one takes the complex above and applies $\operatorname{Hom}_A(-,A)$, the resulting complex is quasi-isomorphic (up to a shift) to the original. In other words, in $\mathrm{D}(A)$ (the derived category of $A$-modules), the Koszul complex $K(\mathbf{x)}$ satisfies $K(\mathbf{x})^\vee \cong K(\mathbf{x})[-\ell]$.
My question is the following:
Does this "self-duality" characterize the Koszul resolution? Namely, suppose $I \subset A$ is an ideal such that $A/I$ has finite projective dimension over $A$. If $Q \to A/I$ is the free resolution of this quotient, such that $Q^\vee \cong Q[-\ell]$, can we conclude that $I$ is generated by a regular sequence of length $\ell$ and $Q$ is its associated Koszul resolution?
I suspect that the answer to the above is no, so my follow up question in that case is if there are any other "categorical" characterizations of when a free resolution is the Koszul resolution of some regular sequence.
Let me give an example of a self-dual free resolution which is not a Koszul complex. Let $$ A = \mathbf{k}[x_{ij}]_{1 \le i < j \le 5} $$ be the polynomial ring in 10 variables and let $M \in \mathrm{Mat}_{5\times 5}(A)$ be the skew-symmetric matrix over $A$ such that $m_{ij} = x_{ij}$ if $i < j$. Then there is an exact sequence $$ 0 \to A \stackrel{\mathrm{Pf}(M)}\to A^{\oplus 5} \stackrel{M}\to A^{\oplus 5} \stackrel{\mathrm{Pf}(M)}\to A \to A/I \to 0, $$ where $\mathrm{Pf}(M)$ is the 5-tuple of Pfaffians of 4-by-4 matrices obtained by removing from $M$ the $i$-th row and column for $1 \le i \le 5$, and $I$ is the ideal generated by these Pfaffians.