If $G$ is a finite group acting on a finite set $X$, we have Burnside's formula that counts the number of orbits $|X/G|$ as: $$ |X/G| = \frac1{|G|} \sum_{g\in G} |X^g|, $$ with $X^g$ being the set of elements in $X$ fixed by $g$.

Now, consider the finite set $X$ of $\mathbb F_p$-points of $\mathbb P^1_{\mathbb F_p}$, the one dimensional projective space over the finite field $\mathbb F_p$, with $p$ elements, and the finite set $Y:=X\times X$ (Cartesian product). The 2-element group $G:=\{e,g\}$ acts by permuting the factors on $Y$ ($G$ is the symmetric group on 2 elements, and $e$ its identity), and it is well-known that (it follows, for example, from the Zeta function computation in https://math.stackexchange.com/q/799101): $$ Y/G = Sym^2(\mathbb P^1_{\mathbb F_p}) = \mathbb P^2_{\mathbb F_p}. $$

As such, we have $|Y/G|=|\mathbb P^2_{\mathbb F_p}|=p^2+p+1$.

On the other hand, one would say that $|Y^e|=|Y|=|X|^2=(p+1)^2$, and $|Y^g|=|X|=p+1$, since the elements fixed under the permutation are in the diagonal of $Y=X\times X$, isomorphic to $X$, so Burnside's formula gives, apparently, the wrong answer: $$ |Y/G|=\frac12((p+1)^2+(p+1))=\frac{p^2+3p+2}{2}.$$

Maybe I am using the wrong notion of $\mathbb F_p$-points of a variety (/scheme?), but I believe there should be a simple explanation of why the computations do not match. Any help is appreciated.