Let $C_{2}$ be the group $\left\langle \tau|\tau^{2}=1\right\rangle $ let $X$ be a finite $C_{2}$-set. Let $\mathbb{Z}^{X}$ be the free $\mathbb{Z}$-module whose generators are $X$. We naturally extend the action of $C_{2}$ on $X$ to an action on $\mathbb{Z}^{X}$. Let $x,y,z\in X$. we define an action of $C_{2}$ on $\mathbb{Z}^{X}\otimes\mathbb{Z}^{X}\otimes\mathbb{Z}^{X}$ by $\tau\left(x\otimes y\otimes z\right)=\tau\left(z\right)\otimes\tau\left(y\right)\otimes\tau\left(x\right)$.
Let $$ \mathbb{Z}^{X}\otimes\mathbb{Z}^{X}\otimes\mathbb{Z}^{X}/\left\langle \tau\left(x\otimes y\otimes z\right)-x\otimes y\otimes z\right\rangle $$ What is its rank?
If $Y$ is a $C_2$-set then $C_2$ acts on the free $\mathbb{Z}$-module $\mathbb{Z}Y$. The quotient map $Y\to Y/C_2$ extends linearly to a module homomorphism $\mathbb{Z}Y\to\mathbb{Z}Y/C_2$ whose kernel is generated by all elements of the form $\sigma y-y$, where $y\in Y$ and $\sigma\in C_2$. Thus, $\mathbb{Z}Y/C_2$ is isomorphic to the module of coinvariants $\mathbb{Z}Y_{C_2}=\mathbb{Z}Y/\langle \sigma y-y\rangle$.
In your case, $\mathbb{Z}X\otimes\mathbb{Z}X\otimes \mathbb{Z}X\cong \mathbb{Z}(X\times X\times X)$, which respects the $C_2$ action, so we are treating the set as $Y=X^3$. To find the rank, we simply need the number of orbits in $X^3$.
Let $n=|X|$ and let $f$ be the number of fixed points of $C_2\curvearrowright X$. Then the number of fixed points in $X^3$ is $f^3$, and the number of other orbits is the number of non-fixed-points halved, i.e. the rank is
$$ f^3+\frac{n^3-f^3}{2}. $$