Isomorphism of A(Zn) and Zn/{0}: A Proof

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Homework Statement



Recall that given a group G, we defined A(G) to be the set of all isomorphisms from G to itself; you proved that A(G) is a group under composition.
(a) Prove that A(Zn) is isomorphic to Zn/{0}
(b) Prove that A(Z) is isomorphic to Z2

Homework Equations


The Attempt at a Solution

 
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There aren't very many automorphisms of Z->Z. In fact, I think there is only two of them. For the other question, I'm not really sure what Z_n/{0} means. Can you explain?
 
Can you figure out the sets A(Zn) and A(Z)?
The identity function is one that should come to mind.
Take a general function on Z by f(x)=bx for some b in Z. What values can b take on so that f(x) is a bijection?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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