For any field $K$, a finite group $G$, and left $G$-set $X$, $KX$ has a obvious left $KG-$ module structure, where $KG$ is the usual group algebra generated by the group $G$.
The question is
Find two non-isomorphic $G$-sets $X$ and $Y$ for which $KX\cong KY$. Especially, do this for $K=\mathbb{C}, G=\mathbb{Z}/{2\mathbb{Z}}$, and $X,Y$ are finite sets.
Restrict to the case $K=\mathbb{C}, G=\mathbb{Z}/{2\mathbb{Z}}$, and $X,Y$ are finite sets.
After some attempts, I think the points are the following:
1, $X\not\cong Y$ as $G$-sets iff they have different orbit structure, since $G$ has two elements, suppose it is generated by $\tau$ so every $G$-set has orbits of two types: orbit with one element $x$, so $\tau(x)=x$, and orbit with two elements $y, \tau(y)$.
2, $KX\cong KY$ may be possible only if $X, Y$ have the same size, and since we can map every element $x_i$ in $X$ to a linear combination of $\sum_{j=1}^mk_{i,j}y_i\in KY$, not necessarily only $y_i$. So we only need to find the possible choices of values on orbit space of $X$ to make sure the matrix the homomorphism determined is invertible.
I have used the above idea (I think they are correct) to construct examples, and I have rulled out the cases $|X|=|Y|=2, 3, 4$, but still got stuck in dealing with higher dimensional cases.
Can anyone give me some hint how to proceed further? or are there any points that I failed to see till now?
Thanks in advance!