Since no-one has explcitly mentioned character theory and representation theory, I will say a few words. If we have a finite group $G$, and two (faithful, ie with trivial kernel) representations $\sigma, \tau : G \to {\rm GL}(n,F)$ where $F$ is a field of characteristic zero, asking whether $G\sigma$ and $G\tau$ are conjugate in ${\rm G}(n,F)$ is the same as asking whether the representations are equivalent over $F.$ An obvious necessary condition is that $g \sigma$ and $g\tau$ have the same trace for all $g \in G,$ but character theory (and some Schur index theory, etc) tell us that this condition is sufficient if the representation is irreducible, and then, with some work, in general (the field $F$ need not be algebraically closed for this conclusion. The (well-known) point is is that in the irreducible case, if the two representations can be intertwined over a n extension of $F,$ they already can be intertwined over $F.$ The question for integral representations is more difficult. It may be that representations $\sigma, \tau : G \to {\rm GL}(n,\mathbb{Z})$ are equivalent after extending the ground ring to $\mathbb{Q},$ but are not equivalent as repesentations over $\mathbb{Z}.$ A relevant theorem putting some control on the situation is one of Jordan-Zassenhaus. An example is that ${\rm GL}(2,\mathbb{Z})$ has two subgroups isomorphi to the dihedral group $D$ with $8$ elements which are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but are conjugate as subgroups of ${\rm GL}(2,\mathbb{Q}).$ The group $D$ had two normal Klein $4$-subgroups $U$ and $V$. If we induce a non-trivial $1$-dimensional representation of $U$ to $D$ we get a subgroup $E$ of ${\rm GL}(2,\mathbb{Z})$ isomorphic to $D,$ and we can to the same for $V$ to get another subgroup $E^{\prime}.$ The subgroups $E,E^{\prime}$ are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but the representations afford the same character, so are equivalent as rational representations. (The technical reason is that if we pass to an appropriate local ring with residue field of characteristic $2,$ we have two indecomposable modules which have non-conjugate vertices, so are not isomorphic).

Since no-one has explicitly mentioned character theory and representation theory, I will say a few words. If we have a finite group $G$, and two (faithful, ie with trivial kernel) representations $\sigma, \tau : G \to {\rm GL}(n,F)$ where $F$ is a field of characteristic zero, asking whether $G\sigma$ and $G\tau$ are conjugate in ${\rm G}(n,F)$ is the same as asking whether the representations are equivalent over $F.$ An obvious necessary condition is that $g \sigma$ and $g\tau$ have the same trace for all $g \in G,$ but character theory (and some Schur index theory, etc.) tell us that this condition is sufficient if the representation is irreducible, and then, with some work, in general (the field $F$ need not be algebraically closed for this conclusion. The (well-known) point is is that in the irreducible case, if the two representations can be intertwined over a n extension of $F,$ they already can be intertwined over $F.$ The question for integral representations is more difficult. It may be that representations $\sigma, \tau : G \to {\rm GL}(n,\mathbb{Z})$ are equivalent after extending the ground ring to $\mathbb{Q},$ but are not equivalent as representations over $\mathbb{Z}.$ A relevant theorem putting some control on the situation is one of Jordan-Zassenhaus. An example is that ${\rm GL}(2,\mathbb{Z})$ has two subgroups isomorphic to the dihedral group $D$ with $8$ elements which are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but are conjugate as subgroups of ${\rm GL}(2,\mathbb{Q}).$ The group $D$ had two normal Klein $4$-subgroups $U$ and $V$. If we induce a non-trivial $1$-dimensional representation of $U$ to $D$ we get a subgroup $E$ of ${\rm GL}(2,\mathbb{Z})$ isomorphic to $D,$ and we can to the same for $V$ to get another subgroup $E^{\prime}.$ The subgroups $E,E^{\prime}$ are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but the representations afford the same character, so are equivalent as rational representations. (The technical reason is that if we pass to an appropriate local ring with residue field of characteristic $2,$ we have two indecomposable modules which have non-conjugate vertices, so are not isomorphic).

Since no-one has explcitly mentioned character theory and representation theory, I will say a few words. If we have a finite group $G$, and two (faithful, ie with trivial kernel) representations $\sigma, \tau : G \to {\rm GL}(n,F)$ where $F$ is a field of characteristic zero, asking whether $G\sigma$ and $G\tau$ are conjugate in ${\rm G}(n,F)$ is the same as asking whether the representations are equivalent over $F.$ An obvious necessary condition is that $g \sigma$ and $g\tau$ hae the same trace for all $g \in G,$ but character theory (and some Schur index theory, etc) tell us that this condition is sufficient if the representation is irreducible, and then, with some work, in general (the field $F$ need not be algebraically closed for this conclusion. The (well-known) point is is that in the irreducible cae, if the two representations can be intertwined over a n extension of $F,$ they already can be intertwined over $F.$ The question for integral representations is more difficult. It may be that representations $\sigma, tau : G \to {\rm GL}(n,\mathbb{Z})$ are eqivalent afte extendng the ground ring to $\mathbb{Q},$ but are not equivalent as repesntations over $\mathbb{Z}.$ A relevant theorem putting some control on the situation is one of Jordan-Zassenhaus. An example is that ${\rm GL}(2,\mathbb{Z})$ has two subroups isomorphi to the dihderal group $D$ wih $8$ ements which are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but are conjugate as subgroups of ${\rm GL}(2,\mathbb{Q}).$ The group $D$ had two normal Klein $4$-subgroups $U$ and $V$. If we induce a non-trivial $1$-dimensional representation of $U$ to $D$ we get a sugroup $E$ of ${\rm GL}(2,\mathbb{Z})$ isomorphic to $D,$ and we can to the same for $V$ to gt another subgroup $E^{\prime}.$ The subgroups $E,E^{\prime}$ are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but the representations afford the same character, so are equivalent as rational representations. (The technical reason is that if we pass to an approriate local ring with residue field of charactristic $2,$ we have two indecomposable modules which have non-conjugate vertices, so are not isomorphic).

Since no-one has explcitly mentioned character theory and representation theory, I will say a few words. If we have a finite group $G$, and two (faithful, ie with trivial kernel) representations $\sigma, \tau : G \to {\rm GL}(n,F)$ where $F$ is a field of characteristic zero, asking whether $G\sigma$ and $G\tau$ are conjugate in ${\rm G}(n,F)$ is the same as asking whether the representations are equivalent over $F.$ An obvious necessary condition is that $g \sigma$ and $g\tau$ have the same trace for all $g \in G,$ but character theory (and some Schur index theory, etc) tell us that this condition is sufficient if the representation is irreducible, and then, with some work, in general (the field $F$ need not be algebraically closed for this conclusion. The (well-known) point is is that in the irreducible case, if the two representations can be intertwined over a n extension of $F,$ they already can be intertwined over $F.$ The question for integral representations is more difficult. It may be that representations $\sigma, \tau : G \to {\rm GL}(n,\mathbb{Z})$ are equivalent after extending the ground ring to $\mathbb{Q},$ but are not equivalent as repesentations over $\mathbb{Z}.$ A relevant theorem putting some control on the situation is one of Jordan-Zassenhaus. An example is that ${\rm GL}(2,\mathbb{Z})$ has two subgroups isomorphi to the dihedral group $D$ with $8$ elements which are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but are conjugate as subgroups of ${\rm GL}(2,\mathbb{Q}).$ The group $D$ had two normal Klein $4$-subgroups $U$ and $V$. If we induce a non-trivial $1$-dimensional representation of $U$ to $D$ we get a subgroup $E$ of ${\rm GL}(2,\mathbb{Z})$ isomorphic to $D,$ and we can to the same for $V$ to get another subgroup $E^{\prime}.$ The subgroups $E,E^{\prime}$ are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but the representations afford the same character, so are equivalent as rational representations. (The technical reason is that if we pass to an appropriate local ring with residue field of characteristic $2,$ we have two indecomposable modules which have non-conjugate vertices, so are not isomorphic).