The paper "Comparison of relative group (co)homologies" suggests there are distinct versions. But I expect the answer to be yes in any sensible version, by mimicking the proof in Brown's bible (Corollary III.8.2).

For example, if K is a normal subgroup of G, then Corollary 4.29 of that "comparison" paper says $H_\ast(G,K)\cong H_\ast(G/K)$. The action should respect this isomorphism, hence trivial.

According to the paper "Comparison of relative group (co)homologies" there are distinct versions (which agree for certain pairs of groups). But I expect the answer to be yes in any sensible version, by mimicking the proof in Brown's bible (Corollary III.8.2).

For example, if K is a normal subgroup of G, then Corollary 4.29 of that "comparison" paper says $H_\ast(G,K)\cong H_\ast(G/K)$. The action should respect this isomorphism, hence trivial.

The paper "Comparison of relative group (co)homologies" suggests there are distinct versions. But I expect the answer to be yes in any sensible version, by mimicking the proof in Brown's bible (Corollary III.8.2). Indeed, I think it already follows from Corollary 4.29 in that "comparison" paper:

Let K be a normal subgroup of G. Then $H_\ast(G,K)\cong H_\ast(G/K)$.

  The action should respect this isomorphism, hence trivial.

The paper "Comparison of relative group (co)homologies" suggests there are distinct versions. But I expect the answer to be yes in any sensible version, by mimicking the proof in Brown's bible (Corollary III.8.2).

For example, if K is a normal subgroup of G, then Corollary 4.29 of that "comparison" paper says $H_\ast(G,K)\cong H_\ast(G/K)$. The action should respect this isomorphism, hence trivial.