Is there a group G such that Aut(G) = $C\_3$? What if we replace 3 with a prime number p?



Aut(G) = $C_3$, G = ?
There is no group $G$ (finite or infinite) for which $Aut(G) \cong C\_p$ (the cyclic group of order $p$), if $p > 1$ is an odd number. 


Suppose otherwise. The inner automorphism group $Inn(G)$ is a subgroup, also cyclic, and a well-known exercise in group theory is that if $Inn(G) \cong G/Z(G)$ is cyclic, then $G$ is abelian. 


An abelian group $G$ has an involution given by inversion. Unless inversion is trivial, we get an element of order 2 in $Aut(G) = C\_p$, contradiction. 


If inversion is trivial, then the abelian group $G$ becomes a vector space over $\mathbb{F}\_2$. In that case it is an easy to prove that either $Aut(G)$ is trivial or has an element of order 2; either way we get a contradiction. 


**Edit:** After listening to some comments about this at meta, I amended my answer so that it gives less away or leaves a bit more to the imagination, or so I hope. 


