I misread your question, so the comments above are answering a different question than you asked. I thought you were asking about the lengths of the sheaf and its dual sheaf, not the lengths of the supports. At any rate, this does follow from duality, since the dual of the dual equals the original module (Proposition 5.1, p. 275 of "Residues and Duality").

Let us work locally: so $R$ is a local, Noetherian, Gorenstein $k$-algebra of dimension $d$ with residue field $k$ (so $k$ is a coefficient field). Denote a dualizing module by $\omega_R \cong R$. Let $E$ be a finite length $R$-module. Let $f\in R$ be any element in the annihilator of $M$. Then since the multiplication by $f$ homomorphism, $$L_{E,f}:E \to E, \ m \mapsto fm,$$ is the zero homomorphism, the same is true on $E^D = \text{Ext}^d_R(E,\omega_R)$, since Ext is an additive functor. Thus, the annihilator ideal of $E$ is contained in the annihilator ideal of $E^D$. On the other hand, $(E^D)^D$ is congruent to $E$ as a finite length $R$-module. Thus, also the annihilator ideal of $E^D$ is contained in the annihilator ideal of $E$. So the two annihilator ideals are equal, i.e., the scheme-theoretic support of $E$ equals the scheme-theoretic support of $E^D$ (although $E$ and $E^D$ need not be isomorphic as $R$-modules).

I misread your question, so the comments above are answering a different question than you asked. I thought you were asking about the lengths of the sheaf and its dual sheaf, not the lengths of the supports. At any rate, this does follow from duality, since the dual of the dual equals the original module (Proposition 5.1, p. 275 of "Residues and Duality").

Let us work locally: so $R$ is a local, Noetherian, Gorenstein $k$-algebra of dimension $d$ with residue field $k$ (so $k$ is a coefficient field). Denote a dualizing module by $\omega_R \cong R$. Let $E$ be a finite length $R$-module. Let $f\in R$ be any element in the annihilator of $M$. Then since the multiplication by $f$ homomorphism, $$L_{E,f}:E \to E, \ m \mapsto fm,$$ is the zero homomorphism, the same is true on $E^D = \text{Ext}^d_R(E,\omega_R)$, since Ext is an additive functor. Thus, the annihilator ideal of $E$ is contained in the annihilator ideal of $E^D$. On the other hand, $(E^D)^D$ is congruent isomorphic to $E$ as a finite length $R$-module. Thus, also the annihilator ideal of $E^D$ is contained in the annihilator ideal of $E$. So the two annihilator ideals are equal, i.e., the scheme-theoretic support of $E$ equals the scheme-theoretic support of $E^D$ (although $E$ and $E^D$ need not be isomorphic as $R$-modules).

I misread your question, so the comments above are answering a different question than you asked. I thought you were asking about the lengths of the sheaf and its dual sheaf, not the lengths of the supports. Here is the easiest counterexample I know for the lengths of the supports. Let $R$ be the local ring $k[x,y]_{\langle x,y \rangle}$, with maximal ideal $\mathfrak{m} = \langle x,y \rangle$. Let $E$ be the finite length module $R/\mathfrak{m}^2 \cong k[x,y]/\langle x^2,xy,y^2 \rangle$. Then the dual $\text{Ext}^2_R(E,\omega_R)$ is isomorphic to the cokernel of the $R$-module homomorphism, $$\phi:R^{\oplus 3} \to R^{\oplus 2}, \ \left[\begin{array}{r} a \\ b \end{array} \right] \mapsto \left[\begin{array}{rr} y & -x & 0 \\ 0 & y & -x \end{array} \right] \left[ \begin{array}{r} a \\ b \end{array} \right].$$ In particular, the support of $E$ is the ideal $\mathfrak{m}^2$ with cokernel $R/\mathfrak{m}^2$, yet the support of $E^D$ is the ideal $\mathfrak{m}$ with cokernel $R/\mathfrak{m}$.

I misread your question, so the comments above are answering a different question than you asked. I thought you were asking about the lengths of the sheaf and its dual sheaf, not the lengths of the supports. At any rate, this does follow from duality, since the dual of the dual equals the original module (Proposition 5.1, p. 275 of "Residues and Duality").

Let us work locally: so $R$ is a local, Noetherian, Gorenstein $k$-algebra of dimension $d$ with residue field $k$ (so $k$ is a coefficient field). Denote a dualizing module by $\omega_R \cong R$. Let $E$ be a finite length $R$-module. Let $f\in R$ be any element in the annihilator of $M$. Then since the multiplication by $f$ homomorphism, $$L_{E,f}:E \to E, \ m \mapsto fm,$$ is the zero homomorphism, the same is true on $E^D = \text{Ext}^d_R(E,\omega_R)$, since Ext is an additive functor. Thus, the annihilator ideal of $E$ is contained in the annihilator ideal of $E^D$. On the other hand, $(E^D)^D$ is congruent to $E$ as a finite length $R$-module. Thus, also the annihilator ideal of $E^D$ is contained in the annihilator ideal of $E$. So the two annihilator ideals are equal, i.e., the scheme-theoretic support of $E$ equals the scheme-theoretic support of $E^D$ (although $E$ and $E^D$ need not be isomorphic as $R$-modules).