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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?

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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 $4$.

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