Is true any group is the automorphism group of itself? I've read not all finite groups are automorphisms of a finite group, but I was reasoning (probably wrongly) like this:
Let's take $\Bbb Z_4$ and try to figure out some automorphisms out if it.
We can define $\Phi(\rho) = \rho^2$, and by this we'll get $\Bbb Z_2$ as automorphism group.
Since the modularity of $\Bbb Z_4$ any power of $g$ which will be $2(1+2n)$ will give the same $\Bbb Z_2$ as result.
Now, if take $\Phi(\rho)=\rho$ i.e. the $\Phi$ as the identity function, we'll have the other automorphism, i.e. $\Bbb Z_4$ itself. This is true for any $1+2n$.
Notice some powers are excluded, like $\Phi(\rho)=\rho^4$, since would send all elements to to the identity, violating the inejctivity of the homomorphism.
So, basically here I see two groups, i.e. $\Bbb Z_2$ and $\Bbb Z_4$.
So, I was guessing: if we take the identity function we will always have the same group as itself automorphism.
So, if this holds, every finite group is the automorphism of a at least one finite group, i.e. itself.
Where am I wrong in my reasoning, please?
Incidentally: I came across this since the book I'm reading defined $\Phi(\rho)=\rho^3$ as the function to be used to compute the $\Bbb Z_4$ automorphism, but using it I stumbled quickly in the whole group, so I realized by myself the right solution is using $2$ as power.
Tx in advance.
 A: No cyclic group, $C_n, n\gt1$, is its own automorphism group,  because generators go to generators, and the automorphism group thus has order $\varphi (n)$.  Since $\varphi (n)\not=n$, the groups are different.
Neither is $\Bbb Z$, which only has $2$ automorphisms.

In fact, this does happen though.  If the center and outer automorphism group are trivial,  this occurs,  and the group is called complete.
For instance,  $S_n, n\not=2,6$.
Just use the isomorphism between $G/Z(G)$ and $\rm{Inn}(G)$.
$S_6$ is not complete,  because it has an outer automorphism.

One more result:  the automorphism group of a non-abelian simple group is complete.
So, for instance,  $A_n,n\ge5, n\ne6$.  In these cases,  $\rm{Aut}(A_n)\cong S_n$.
$\rm{Aut}(A_6)\cong P\Gamma L(2,9)$.  It's the "projective semi-linear group of degree $2$ over the field of $9$ elements".   And it's complete.

There are groups that are not complete,  but still equal to their automorphism group.   For instance $D_8$.  It isn't complete because it's center is $\Bbb Z_2$.  (It also has outer automorphisms.  All the dihedral groups do except $D_6$.)
