smc

Setting the first two equations equal to each other, ${\log }_{3}x={\log }_{x}3$. Solving this, we get $\left(3,1\right)$ and $\left(\frac{1}{3},-1\right)$. Similarly with the last two equations, we get $\left(3,-1\right)$ and $\left(\frac{1}{3},1\right)$. Now, by setting the first and third equations equal to each other, we get $\left(1,0\right)$. Pairing the first and fourth or second and third equations won't work because then $\log x\le 0$. Pairing the second and fourth equations will yield $x=1$, but since you can't divide by $\log 1=0$, it doesn't work. After trying all pairs, we have a total of $5$ solutions $\to \boxed{5}$