Is a cubic hypersurface determined by its Fano variety of lines?

Edit. Regarding the Picard group, since $\rho^{-1}(F(X))$ is a $\mathbb{P}^1$-bundle over $F(X)$, the Picard group $\text{Pic}(\rho^{-1}(F(X)))$ equals $\text{Pic}(F(X))\oplus \mathbb{Z}\cdot [\pi^*\mathcal{O}(1)]$. The fibers of the projection, $\pi:\rho^{-1}(F(X))\to X$, are complete intersections of dimension $n-3$ over $X^o$. Thus, for $n\geq 6$, by the Grothendieck-Lefschetz theorem on Picard groups from SGA 2, also $\text{Pic}(\pi^{-1}(X^o)\cap \rho^{-1}(F(X)))$ equals $\mathbb{Z}\cdot [\rho^*\mathcal{O}(1)] \oplus \text{Pic}(X^o)$. Since the complement of $X^o$ in $X$ has codimension $\geq 2$, also $\text{Pic}(X^o)$ equals $\text{Pic}(X)$, and this equals $\mathbb{Z}\cdot[\mathcal{O}(1)|_X]$. If $n$ equals $5$, consider the threefolds $\pi^{-1}(\ell)\cap \rho^{-1}(F(X))$, where $\ell\subset X$ is a general line. Intersecting with a general hyperplane $H$ gives a birational version of this threefold in the cubic fivefold $X$, and now I believe we can repeat the argument from the Griffiths-Harris proof of the Noether-Lefschetz theorem to conclude that $\pi^{-1}(\ell)\cap \rho^{-1}(F(X))$ has Picard rank $2$ (this involves analyzing the singularities of the birational version of the threefold).

Original post. I am adding some details to my comment, specifically the proof that the restriction homomorphism is injective, $$H^0(\text{Grass}(\mathbb{P}^1,\mathbb{P}^{n+1}),\mathcal{O}(2)) \to H^0(F(X),\mathcal{O}(2)).$$ One proof uses the flag variety $$\text{Flag}_{0,1}(\mathbb{P}^{n+1}) \subset \mathbb{P}^{n+1}\times_{\text{Spec}\ k} \text{Grass}(\mathbb{P}^1,\mathbb{P}^{n+1}),$$ with its two projections, $$\pi:\text{Flag}_{0,1}(\mathbb{P}^{n+1}) \to \mathbb{P}^{n+1}, \ \ \rho:\text{Flag}_{0,1}(\mathbb{P}^{n+1}) \to \text{Grass}(\mathbb{P}^1,\mathbb{P}^{n+1}).$$ For a hypersurface $X\subset \mathbb{P}^{n+1}$, for the inverse image $\pi^{-1}(X)$, the induced morphism is dominant, $$\rho:\pi^{-1}(X)\to \text{Grass}(\mathbb{P}^1,\mathbb{P}^{n+1}),$$ and it is even flat over the open complement of the Fano scheme $F(X) = \text{Hilb}^{t+1}_{X/k}.$ Thus, the following pullback map is injective, $$H^0(\text{Grass}(\mathbb{P}^1,\mathbb{P}^{n+1}),\mathcal{O}(2)) \to H^0(\pi^{-1}(X), \rho^*\mathcal{O}(2)).$$ Thus, it suffices to prove that the following pullback map is injective, $$H^0(\text{Grass}(\mathbb{P}^1,\mathbb{P}^{n+1}),\mathcal{O}(2)) \to H^0(\pi^{-1}(X)\cap \rho^{-1}(F(X)),\rho^*\mathcal{O}(2)).$$ There is a maximal open subscheme $X^o\subset X$ over which the following morphism is flat of relative dimension $\leq n-d$ (in fact dimension precisely $n-d$), $$\pi:\pi^{-1}(X)\cap \rho^{-1}(F(X)) \to X.$$ It suffices to prove injectivity of the map, $$H^0(\text{Grass}(\mathbb{P}^1,\mathbb{P}^{n+1}),\mathcal{O}(2)) \to H^0(\pi^{-1}(X^o)\cap \rho^{-1}(F(X)),\rho^*\mathcal{O}(2)).$$ For $n\geq 3$, the Fano scheme $F(X)$ is smooth of dimension $2n-4$ and $X_o$ is a dense open subscheme whose complement has codimension $\geq 2$ everywhere.

There is a sequence of Cartier divisors, $$X^o_0 = \pi^{-1}(X^o)\supseteq X^o_1 \supseteq X^o_2 \supseteq X^o_3 = \pi^{-1}(X^o)\cap \rho^{-1}(F(X)),$$ where the Cartier divisor $X^o_{e+1}$ in $X^o_e$ is the zero scheme of a section $s_{e+1}$ of the restriction to $X^o_e$ of the invertible sheaf, $$L_{3-(e+1),e+1} = \pi^*\mathcal{O}(3-(e+1))\otimes \rho^*\mathcal{O}(e+1).$$ For the ideal sheaf $\mathcal{I}$ of $\pi^{-1}(X^o)\cap \rho^{-1}(F(X))$ inside $\pi^{-1}(X^o)$, the pushforward locally free sheaf $$I_2:=\pi_*(\mathcal{I}\otimes \rho^*\mathcal{O}(2))$$ has an associated filtration that gives a short exact sequence, $$ 0\to J_2 \to I_2 \to \mathcal{O}(-(3-2))|_{X^o}, $$ where $J_2$ is the pushforward of the twist by $\rho^*\mathcal{O}(2)$ of the ideal sheaf $\mathcal{J}_1$ of $X_1^o$. The scheme $X_1^o$ parameterizes pointed lines that are tangent to $X$ at the marked point. The image of $X_1^o$ under $\rho$ is a dense open in a hypersurface in the Grassmannian, and the degree of that hypersurface equals $6$ (the degree of the dual curve of a plane cubic). Since $6$ is greater than $2$, there are no global sections of $J_2$. Also, there are no global sections of $\mathcal{O}(-1)|_{X^o}$. Thus, there are no global sections of $I_2$. Therefore, the only section of $\mathcal{O}(2)$ on $\text{Grass}(\mathbb{P}^1,\mathbb{P}^{n+1})$ that vanishes on $F(X)$ is the zero section.

For what it's worth, I think I now know a complete proof. I learned of this proof from D. Huybrechts's notes in progress http://www.math.uni-bonn.de/people/huybrech/Notes.pdf, proposition 6.21.

Theorem. Let $X$ be a cubic hypersurface of dimension $d\geqslant 3$, with $d\neq 4$. Then $X$ can be recovered from its Fano variety of lines $F(X)$.

Notations: $F=F(X)$, $G=\mathrm{Gr}(2,d+2)$, $\mathscr{O}_F(1)$ and $\mathscr{O}_G(1)$ the Plücker polarisations of $F$ and $G$, respectively.

For future reference, let me state the following results.

Lemma 1. The canonical map $\mathrm{H}^0(G,\mathscr{O}_G(2))\rightarrow\mathrm{H}^0(F,\mathscr{O}_F(2))$ is injective.

Proof. See J. Starr's answer or Altman-Kleiman, Foundations of the theory of Fano schemes, proposition 1.15 (ii).

Lemma 2. $\omega_{F}=\mathscr{O}_{F}(4-d)$.

Proof. See Altman-Kleiman, proposition 1.8.

Lemma 3. If $V$ is a variety such that $\omega_{V}^{\vee}$ is ample, then $\mathrm{Pic}(V)$ is torsion-free.

Proof. Let $L$ be a torsion line bundle on $V$. Then $\mathrm{H}^i(V,L)=0=\mathrm{H}^i(V,\mathscr{O}_V)$ for $i\geqslant 1$ by Kodaira vanishing. As $\chi(L)=\chi(\mathscr{O}_V)$ by Riemann-Roch, we get $\mathrm{H}^0(V,L)=\mathrm{H}^0(V,\mathscr{O}_V)$, which forces $L$ to be trivial.

Proof of the Theorem. Let $X$ and $X'$ be cubic hypersurfaces and $f:F\xrightarrow{\sim} F'$ an isomorphism between their Fano varieties of lines. By lemma 2 we have the equality $$(4-d)f^\ast(\mathscr{O}_{F'}(1))=f^\ast(\omega_{F'})=\omega_F=(4-d)\mathscr{O}_{F}(1)$$ in $\mathrm{Pic}(F)$. For $d\geqslant 5$ lemma 2 shows that we can use lemma 3 to deduce $$f^\ast(\mathscr{O}_{F'}(1))=\mathscr{O}_{F}(1).$$ For $d=3$ this is clear. The theorem then follows from proposition 4 of https://arxiv.org/abs/1209.4509. For completeness, I repeat the proof. Let $\mathbf{P}^{d+1}=\mathbf{P}(V)$ be the ambient projective space containing $X$ and $X'$. By lemma 1, if $Q$ is a quadric in $\mathbf{P}(\wedge^2 V)$ containing $F$, then $Q$ also contains $G$. In $\mathbf{P}(\wedge^2 V)$ the Grassmannian $G$ is cut out by quadrics, so in particular $G$ is the intersection of quadrics containing $F$. Hence there is an automorphism $g$ of $G$ taking $F$ to $F'$ via $f$, and $g$ also preserves the Plücker polarisation. By a theorem of Chow on automophisms of Grassmannians (Ann. Math. 2), $g$ is induced by an automorphism $h$ of $\mathbf{P}(V)$. The latter automorphism then induces a (polarisation-preserving) isomorphism $X\xrightarrow{\sim} X'$. (If $L$ is a line on $X$ corresponding to $[L]\in F$, then $h(L)$ is a line on $X'$, and $[h(L)]=f([L])$. Of course $X$ is covered by lines.)