For equivalence of unbranched coverings of topological spaces, there is a criteria:
Two coverings (unbranched) $p_1\colon Y_1\rightarrow X$ and $p_2\colon Y_2\rightarrow X$ are equivalent iff for some $q\in X$ and $\bar{q_1}\in p_1^{-1}(q)$ and $\bar{q_2}\in p_2^{-1}(q)$, the induced subgroups $p_*\pi_1(Y_1,\bar{q_1})$ and $p_*\pi_1(Y_2,\bar{q_2})$ are conjugate in $\pi_1(X,q)$.
Is there any criteria (particularly group theoretic) for equivalence of branched coverings of Riemann surfaces?
The answer is yes: the equivalence class of the covering is detected by the monodromy representation of the fundamental group of the base minus the branch locus, up to conjugacy.
More precisely, let $f \colon X \to Y$ be a (possibly branched) covering of degree $d$ of Riemann surfaces. Choosing a point $y_0 \in Y$ not lying in the branch locus $B$, there is a monodromy representation $$\rho \colon \pi_1(Y-B, y_0) \to S_d \quad (*)$$ whose image is transitive (since $X$ is assumed to be connected). Moreover, if we choose a different base point it is easy to check that the map $\rho$ varies only up to conjugacy in $S_d$.
Then there is the following well-known
Theorem. There exists a one-to-one correspondence between branched coverings $f \colon X \to Y$ of degree $d$ whose branch points lie in $B$ and group homomorphisms of type $(*)$ with transitive image (up to conjugacy in $S_d$).
For further details, see Miranda's book Algebraic curves and Riemann surfaces, especially Chapter 4.
