Image of square under Mobius Transformation

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Am i right that there is no Mobius transformation $$ h(z) $$ that sends a square to rectangle "A" with vertices in $$ 0, 2, i, i+2 $$because if we inscribe a cricle into square then it has 4 points $$ z_1,\ z_2,\ z_3,\ z_4 $$on a boundry on square therefore $$ h(z_1), \ h(z_2), \ h(z_3), \ h(z_4)$$ also belongs to boundry of image of square under h and image of square is generlised circle whih is contradiction because we can't construct a cricle through theses points?