Example C_n^2:C_3
Group $G=C_n^2:C_3$ is described by the short exact sequence
\[\begin{align*} 1 \to C_n \oplus C_n \to G \to C_3 \to 1 \end{align*}\]When $n=5$, group $G=C_5^2:C_3 \subseteq \rm{PGL}(3,\mathbb{C})$, it has 2 projective linear representations $\mathbb{P}(V_1),\mathbb{P}(V_2)$ where
\[\begin{align*} G= \langle g_{n,s}^1=\begin{pmatrix} \zeta_n^s & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}, g_{n,t}^2=\begin{pmatrix} 1 & 0 & 0\\ 0 & \zeta_n^t & 0\\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix} \rangle \subseteq \rm{PGL}(3,\mathbb{C}). \end{align*}\]where for $V_1$, we choose $s=1,t=2$ and for $V_2$, we choose $s’=3,t’=4$.
As a linear group in $\rm{GL}(3,\mathbb{C})$, it has center
\[\begin{align*} C_n= \langle \begin{pmatrix} \zeta_n & 0 & 0\\ 0 & \zeta_n & 0\\ 0 & 0 & \zeta_n \end{pmatrix} \rangle. \end{align*}\]The Magma code is as follows:
n:=5; FScale:=CyclotomicField(3:Sparse:=true); FScale:=Compositum(FScale,CyclotomicField(n^2:Sparse:=true)); F:=FScale; Zn:=RootOfUnity(n); s1:=1; t1:=2; G1:=MatrixGroup<3,F| [Zn^(s1),0,0, 0,1,0, 0,0,1], [1,0,0, 0,Zn^(t1),0, 0,0,1], [0,0,1, 1,0,0, 0,1,0]>; s2:=3; t2:=4; G2:=MatrixGroup<3,F| [Zn^(s2),0,0, 0,1,0, 0,0,1], [1,0,0, 0,Zn^(t2),0, 0,0,1], [0,0,1, 1,0,0, 0,1,0]>;
The Burnside Symbols are
\[\begin{align*} [\mathbb{P}(V_1) \circlearrowleft G]=&(1,G \circlearrowright k(x,y),())\\ &+(C_5,C_5^2/C_5 \circlearrowright k(x),(1))+(C_5,C_5^2/C_5 \circlearrowright k(x),(4))\\ &+(C_5^2,1 \circlearrowright k,((1,4),(3,1)))+(C_5^2,1 \circlearrowright k,((4,0),(4,1))) \end{align*}\]where
\[\begin{align*} C_5=\langle \begin{pmatrix} \zeta_5^4 & 0 & 0\\ 0 & \zeta_5^4 & 0\\ 0 & 0 & 1 \end{pmatrix} \rangle, C_5^2=\langle \begin{pmatrix} \zeta_5^3 & 0 & 0\\ 0 & \zeta_5^4 & 0\\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix} \zeta_5 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \rangle \end{align*}\]There are conjugation relations:
\[\begin{align*} &(C_5^2,1 \circlearrowright k,((1,4),(3,1)))=(C_5^2,1 \circlearrowright k,((1,0),(1,4)))=(C_5^2,1 \circlearrowright k,((1,0),(3,1))),\\ &(C_5^2,1 \circlearrowright k,((4,0),(4,1)))=(C_5^2,1 \circlearrowright k,((2,4),(4,0)))=(C_5^2,1 \circlearrowright k,((2,4),(4,1))). \end{align*}\] \[\begin{align*} [\mathbb{P}(V_2) \circlearrowleft G]=&(1,G \circlearrowright k(x,y),())\\ &+(C_5,C_5^2/C_5 \circlearrowright k(x),(2))+(C_5,C_5^2/C_5 \circlearrowright k(x),(3))\\ &+(C_5^2,1 \circlearrowright k,((3,3),(4,2)))+(C_5^2,1 \circlearrowright k,((2,0),(2,2))) \end{align*}\]where
\[\begin{align*} C_5=\langle \begin{pmatrix} 1 & 0 & 0\\ 0 & \zeta_5^2 & 0\\ 0 & 0 & 1 \end{pmatrix} \rangle, C_5^2=\langle \begin{pmatrix} 1 & 0 & 0\\ 0 & \zeta_5^3 & 0\\ 0 & 0 & \zeta_5 \end{pmatrix}, \begin{pmatrix} \zeta_5^2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \rangle \end{align*}\]There are conjugation relations:
\[\begin{align*} &(C_5^2,1 \circlearrowright k,((3,3),(4,2)))=(C_5^2,1 \circlearrowright k,((3,0),(3,3)))=(C_5^2,1 \circlearrowright k,((3,0),(4,2))),\\ &(C_5^2,1 \circlearrowright k,((2,0),(2,2)))=(C_5^2,1 \circlearrowright k,((1,3),(2,0)))=(C_5^2,1 \circlearrowright k,((1,3),(2,2))). \end{align*}\]Now if we fix generator for $C_5=\langle (g_{n,s}^1)^t(g_{n,t}^2)^s\rangle$ and $C_5^2=\langle g_{n,s}^1,g_{n,t}^2\rangle$. Two Burnside Symbols are
\[\begin{align*} [\mathbb{P}(V_1) \circlearrowleft G]=&(1,G \circlearrowright k(x,y),())\\ &+(C_5,C_5^2/C_5 \circlearrowright k(x),(2))+(C_5,C_5^2/C_5 \circlearrowright k(x),(3))\\ &+(C_5^2,1 \circlearrowright k,((0,3),(1,0)))+(C_5^2,1 \circlearrowright k,((0,2),(4,0))) \end{align*}\] \[\begin{align*} [\mathbb{P}(V_2) \circlearrowleft G]=&(1,G \circlearrowright k(x,y),())\\ &+(C_5,C_5^2/C_5 \circlearrowright k(x),(2))+(C_5,C_5^2/C_5 \circlearrowright k(x),(3))\\ &+(C_5^2,1 \circlearrowright k,((0,1),(3,0)))+(C_5^2,1 \circlearrowright k,((0,4),(2,0))) \end{align*}\]The combinatorial Burnside group of $C_5^2:C_3$ is \(\begin{align*} \mathcal{BC}_2(C_5^2:C_3)=(\mathbb{Z}/2)^2 \times (\mathbb{Z}/30)^2 \times \mathbb{Z}^{19} \end{align*}\)
Remark: we don’t consider symbols with trivial stabilizer group in the computation of $\mathcal{BC}$.
Let $T_2^{1,2},T_{30}^{1,2}$ and $e_i$, $i=1,\dots,19$ be the generators for $(\mathbb{Z}/2)^2$, $(\mathbb{Z}/30)^2$ and the torsion-free part respectively.
The difference of images of two Burnside symbols $[\mathbb{P}(V_1) \circlearrowleft G]-[\mathbb{P}(V_2) \circlearrowleft G]$ under the map $\rm{Burn}_2(G) \to \mathcal{BC}_2(G)$ is:
\[\begin{align*} T_2^2+13T_{30}^2+e_3-e_7+e_8+2e_{11}-e_{12}+e_{16}\neq 0 \end{align*}\]Thus two actions are not birational to each other.