Some general remarks on your question: there is an algorithm which, for any $g$, will return a list of all automorphisms of a genus $g$ surface. I'll try to give one possible interpretation of what "list" might mean.
By the Hurwitz estimate, one knows that any automorphism group of a genus $g$ surface is of order $\leq 84(g-1)$. Moreover, if this bound is sharp, then the quotient orbifold of the automorphism group is the $(2,3,7)$ triangle orbifold. To determine when this is realized, you may take the $(2,3,7)$ group, and try to find epimorphisms to finite groups of size $84(g-1)$ for which every (non-identity) torsion element is mapped non-trivially. The cover corresponding to the kernel of such an epimorphism is a surface of genus $g$. Larsen has used this observation to estimate how often Hurwitz's bound is achieved. In principle, for any given $g$, one could try to classify all groups of order $84(g-1)$ and then count epimomorphisms to these groups from the $(2,3,7)$ group by computing where the generators are sent. There are only finitely many conjugacy classes of torsion elements, so one may compute whether these epimorphisms map non-trivially on torsion. Two such epimorphisms will give the same cover if and only if the kernels are the same, which is again (in principle) computable (this follows because the $(2,3,7)$ triangle groups has no non-trivial outer automorphisms).
More generally, suppose one chooses an $a\leq 84(g-1)$. Then any group of order $a$ acting on a surface of genus $g$ gives a quotient orbifold of Euler characteristic $(2-2g)/a$. One may compute all orbifolds of this Euler characteristic, compute the fundamental groups, and compute all epimorphisms to groups of size $a$ (which are non-trivial on torsion). The automorphism groups acting on the genus $g$ surface associated to such epimorphisms of the quotient orbifold will be equivalent if and only if the the kernels of the epimorphisms are equivalent by an automorphism of the orbifold fundamental group (since inner automorphisms fix normal subgroups, one need only consider the outer automorphism group). This is in principle computable. There are finitely many triangulations of the quotient orbifold up to the action of the mapping class group, with vertices on the singular points of the orbifold (or one-vertex triangulations if the action is free). One may compute all possible such triangulations, up to combinatorial equivalence (the action of the mapping class group).
Two epimorphisms of the same orbifold give rise to the same $g$ action if and only if the kernels of these epimorphisms are in the same orbit of the mapping class group action, which will occur if the triangulations agree for some choice of triangulation. One may navigate through all triangulations by doing Whitehead moves, to determine if two triangulations are equivalent.
So one possible answer to your question would be a list of orbifolds of Euler characteristic $(2-2g)/a$, $a\leq 84(g-1)$, together with epimorphisms of the orbifold fundamental groups to a group of order $a$ which are non-trivial on torsion. Using the described algorithms, one could ensure that this list had precisely one representative of each automorphism group. One could also try to express these as automorphisms of the surface by choosing a triangulation for the orbifold, and using this to get a tiling of the covering space together with the action of the group on this tiling. But I do not know of a canonical way to represent the automorphism group on a triangulation, unless the quotient is a nice rigid orbifold like a triangle or turnover orbifold.