Find All Automorphisms of Cyclic Group of Order 10

1. The problem statement, all variables and given/known data

Find all the automorphisms of a cyclic group of order 10.

2. Relevant equations

ψ(a)ψ(b)=ψ(ab)

For G= { 1, x, x^2,..., x^9}, and some function

ψ(a) = x^(a/10)

3. The attempt at a solution

I know that a homomorphism takes the form

Phi(a)*phi(b) = phi (ab) , and that an automorphism maps from G->G,

However, I don't understand what an automorphism for a cyclic group would even look like. I suppose it should be something of the form:

ψ(a) = x^(a/10)

and that a should be a specific power, but I have no idea where to go from here.

I appreciate any help. Thanks

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 Recognitions: Homework Help Science Advisor Think about it. An automorphism has to map a generator of the cyclic group to another generator. How many elements of your cyclic group are generators?
 I don't have a solid understanding of generators. I understand that you need relatively prime powers to map out all the functions through multiplication. Going by this, I suppose the generators would be 1,3,5,7, & 9, all the odd numbers below 10. For what reason are would these be automorphisms though? Tell me if this is right: [ If x^n is an automorphism, where n is an individual element of {1,3,5,7,9}, the function works by multiplying each term of G= {1,x,...,x^9} so that phi(a) = x^n *x*a = x^ (n+a) It's a homomorphism because: phi(a) * phi (b) = x^(n+a) * x^(n+b) = x^( 2n+a+b) and phi(ab) = x^(n + a +b) (I know 2n+a+b ≠ n+a+b, but it's the best I can come up with) Then the function is bijective because if phi(x) = x^n, phi(x)^-1 = x^-n or x^(10-x) because of the cyclic nature Because it maps from G->G, it is homomorphic, and has an inverse, it is bijective automorphism for the cyclic group of order 10. ] Am I at least heading in the right direction with this?

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