smc

We have $ab=c$, $bc=a$, and $ca=b$, so multiplying these three equations together gives $a^2{b}^2{c}^2=abc⟹abc(abc-1)=0$, and as $a$, $b$, and $c$ are all non-zero, we cannot have $abc=0$, so we must have $abc=1$. Now substituting $bc=a$ gives $a(bc)=1⟹a^2=1⟹a=\pm 1$. If $a=1$, then the system becomes $b=c,bc=1,c=b$, so either $b=c=1$ or $b=c=-1$, giving $2$ solutions. If $a=-1$, the system becomes $-b=c,bc=-1,-c=b$, so $-b=c=1$ or $b=-c=1$, giving another $2$ solutions. Thus the total number of solutions is $2+2=4$, which is answer $\boxed{4}$.