What are the fixed points of the jacobian acting on the compactified jacobian ? Let C be an integral projective curve over $\mathbb{C}$ and Jac(C) be its jacobian.
Let $\overline{Jac(C)}$ be the compactified jacobian of C (the moduli space of rank 1 torsion free
sheaves of degree 0 on C). Jac(C) acts on $\overline{Jac(C)}$ by tensor product.
The question is :
What are the fixed points of this action ?
$\overline{Jac(C)}$ contains Jac(C) and obviously there is no fixed point on Jac(C) because
the restriction to the action to it is a action by translation. So the fixed points are on 
$\overline{Jac(C)}$ minus Jac(C).
If C is rational (of geometric genus 0) and if $f: \mathbb{P}^{1} \longrightarrow C$ is the normalisation, there is a natural point in $\overline{Jac(C)}$ : F = $f_{*} \mathcal{O}_{\mathbb{P}^{1}}$. A more restricted question is :
Is F always a fixed point of the action ? Is it the once ?
(it is true for a nodal or cuspidal C but surely these examples are too simple ...)
 A: I think that's the only fixed point.


*

*The compactified Jacobian is a homeomorphic to a product of the Jacobian of the normalization times some local factors from the singularities; likewise the Jacobian splits into the Jacobian of the normalization times some local factors.  In particular this means that there are certainly no fixed points if the geometric genus is > 0, since the Jacobian of the normalization acts freely; also the splitting into the product of the contributions for the singularities means you might as well consider a curve with a unique singularity.

*If $u:C' \to C$ is the minimal unibranch partial normalization (i.e. you separate all the branches but don't do anything else), then I believe the entire fixed locus lies in the pushforward of torsion free sheaves on $C'$; you should find a proof in Beauville's article on rational curves on K3 surfaces.  In particular if the curve is immersed, there are no other fixed points than the one you describe.

*Certainly if the normalization is $n:\mathbb{P}^1 \to C$, then $n_* \mathcal{O}$ is a fixed point.  By the projection formula, $L \otimes n_*(\mathcal{O}) = n_{*} (n^*L \otimes \mathcal{O}) = n_{*}(\mathcal{O})$.

*Passing to the complete local ring $R$ at the (now unibranch) singularity, the space of torsion free sheaves (of some fixed degree) is constructibly (with respect to the orbits of the Jacobian) in bijection with the space of $R$-modules $R'$ with $\mathbb{C}[[t]] > R' > \mathrm{Conductor}(R)$, with the same relative dimension as $R$.  The action of the Jacobian is just the action of the invertible power series (modulo the action of $R^*$).  From this it is clear the only fixed point is some $t^k \mathbb{C}[[t]]$ itself, i.e. the pushforward of the normalization.
p.s. I may have assumed characteristic zero at some point.
