I think that's the only fixed point. Here's a probably unnecessarily complicated argument. I may have restricted to characteristic zero at some point.

1. 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.

2. 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.

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

4. 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 $R$-submodules $R'$ with $\mathbb{C}[[t]] > R' > R$, and 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 $\mathbb{C}[[t]]$ itself, i.e. the pushforward of the normalization.

I think that's the only fixed point.

1. 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.

2. 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.

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

4. 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.