Find a group $H$ of permutations such that $U(10) \cong H$
My idea:
since $U(1)=\{1,3,7,9\}$
since order of 1 is 1
order of 3 is 4
order of 7 is 4
order of 9 is 2
so we have to find a permutation H with four elements and order of these elements are equal order of the elements of U(10)
is this idea is right