jmc

If $x=2$ or $x=4,$ then the fraction is undefined. Otherwise, we can cancel the factors of $(x-2)(x-4)(x-2),$ to get $$(x-1)(x-3)(x-3)(x-1)=1.$$Then $(x-1{)}^2(x-3{)}^2-1=0,$ so $[(x-1)(x-3)+1][(x-1)(x-3)-1]=0.$

If $(x-1)(x-3)+1=0,$ then $x^2-4x+4=(x-2{)}^2=0.$ We have already ruled out $x=2.$

If $(x-1)(x-3)-1=0,$ then $x^2-4x+2=0.$ By the quadratic formula, $$x=2\pm \sqrt{2}.$$Thus, the solutions are $\boxed{2+\sqrt{2},2-\sqrt{2}}.$