The transitive simultaneous conjugacy problem asks whether there exists a permutation such that holds for all , where and are given sequences of permutations in , each of which generates a transitive subgroup of . As from mid 70’ it has been known that the problem can be solved in time. An algorithm with running time $O(dn \log(dn))$, proposed in late 80’, does not work correctly on all input data. In this paper we solve the transitive simultaneous conjugacy problem in $O(n^2 \log d / \log n + dn\log n)O(n^{3/ 2} + dn)$ space. Experimental evaluation on random instances shows that the expected running time of our algorithm is considerably better, perhaps even nearly linear in at given .

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