Let $D$ be the normalization of $C$. The map $D \to C$ comes from a canonical construction, so automorphisms of $C$ lift uniquely to automorphisms of $D$.

So $Aut(C) \subseteq PGL_2$.

Next note that $Aut(C)$ is contained in the group of automorphism of $D$ that fix the inverse image of the singular points of $C$. So if this set has size at least $3$, the group is finite.

There are not too many subgroups of $PGL_2$. For instance, the finite ones are all either $C_n$, $D_n$, $A_4$, $S_4$, or $A_5$. To construct a curve whose automorphism group is one of these, any easy method is to take $\mathbb P^1$, view it as a sphere, and glue all the vertices of the appropriate regular polyhedron together.

The infinite ones must have connected component of the identity $\mathbb G_m$, $\mathbb G_a$, or $\mathbb G_a \ltimes \mathbb G_m$. I believe the only infinite ones are these, $\mathbb G_m \ltimes C_2$, and $\mathbb G_a \ltimes C_n$. I think it is not too hard to construct examples of all of these.

This leaves the question of when a rational curve with one or two singular points that lift to one or two points in the normalization has an infinite automorphism group. The answer will come down to local obstructions at the singular point(s) - does scaling induce a symmetry of the singularity? Does the operation $x \to \frac{1}{\frac{1}{x}+t}$ induce a symmetry? I don't know a nice way of checking which singularities have this symmetry.