I don't think this is what you meant to say (it is trivially true for any element of a group).
The question is asking you to find
a∈Z10∗ such that for any
x∈Z10∗ we can find an integer n with
an=x.
Much less formally, it's asking you to find a member of
Z10∗ where if we take successive powers, we get
every element of
Z10∗.
I won't answer the question for you, but suppose we were looking at
Z14∗ instead. This has elements 1, 3, 5, 9, 11, 13 (we exclude multiples of 2 and 7).
We simply try the elements in turn until we find one that works.
Taking powers of 1 we get 1, 1, 1, 1, etc. which obviously doesn't give us every element.
Taking powers of 3 gives 3, 9, 27 = 13 mod 14, 39 = 11 mod 14, 33 = 5 mod 14, 15 = 1 mod 14.
So powers of 3 gives us 3, 9, 13, 11, 5, 1, which does give us all the elements.
So 3 would be a solution if we were working in
Z14∗.
You need to do something similar in
Z10∗.