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From the given conditions, $|{\alpha }^2-1|=2|\alpha -1|$ and $|{\alpha }^4-1|=4|\alpha -1|.$ From the first equation, $$|\alpha +1||\alpha -1|=2|\alpha -1|.$$Since $\alpha \ne 1,$ $|\alpha -1|\ne 0.$ Thus, we can safely cancel the factors of $|\alpha -1|,$ to get $$|\alpha +1|=2.$$From the second equation, $$|{\alpha }^2+1||{\alpha }^2-1|=4|\alpha -1|.$$Then $2|{\alpha }^2+1||\alpha -1|=4|\alpha -1|,$ so $$|{\alpha }^2+1|=2.$$Let $\alpha =x+yi,$ where $x$ and $y$ are real numbers. Then ${\alpha }^2=x^2+2xyi-y^2,$ so the equations $|\alpha +1|=2$ and $|{\alpha }^2+1|=2$ becomes $$\begin{array}{rl}|x+yi+1|& =2,\\ |x^2+2xyi-y^2+1|& =2.\end{array}$$Hence, $$\begin{array}{rl}(x+1{)}^2+y^2& =4,\\ (x^2-y^2+1{)}^2+(2xy{)}^2& =4.\end{array}$$From the first equation, $y^2=4-(x+1{)}^2=3-2x-x^2.$ Substituting into the second equation, we get $$(x^2-(3-2x-x^2)+1{)}^2+4x^2(3-2x-x^2)=4.$$This simplifies to $8x^2-8x=0,$ which factors as $8x(x-1)=0.$ Hence, $x=0$ or $x=1.$

If $x=0,$ then $y^2=3,$ so $y=\pm \sqrt{3}.$

If $x=1,$ then $y^2=0,$ so $y=0.$ But this leads to $\alpha =1,$ which is not allowed.

Therefore, the possible values of $\alpha$ are $\boxed{i\sqrt{3},-i\sqrt{3}}.$

Alternative: We can rewrite the second equation as $(x^2+y^2+1{)}^2-4y^2=4.$ From the first equation, we have $x^2+y^2+1=4-2x$ and $y^2=4-(x+1{)}^2.$ Substituting these, we get $$(4-2x{)}^2-4(4-(x+1{)}^2)=4.$$This simplifies to $8x^2-8x=0,$ and we can continue as before.