Let $G$ be finite group and order of $Aut(G)$ is prime number $p$ then prove that order of $G$ is always less than equal to $3$? As order of $Aut(G)$ is prime no. Then this implies $Aut(G)$ is cyclic this means $Aut(G)$ is abelian this implies inner automorphism group is also cyclic, as cyclic subgroup of cyclic group is cyclic hence as $Inn(G)$ is isomorphic to $G/Z(G)$.
And as $G/Z(G)$ is cyclic therefore $G$ is abelian. 
Then how can we say group order cannot exceed $3$. 
 A: Some hints: since $G$ must be abelian, the map $g \mapsto g^{-1}$ gives rise to an automorphism of order $2$. Hence either this map is the trivial one, that is $g^2=1$ for all $g \in G$, and $G$ must be a direct product of copies of $C_2$. Or $p=2$. Can you finish?
A: We can't.

So far, you know that  $G$ is a finite abelian group, so is the direct sum $$\tag1G=\bigoplus_i C_i$$ of cyclic groups $$C_i=\Bbb Z/\Bbb p_i^{a_i}\Bbb Z$$  of prime power order ($a_i\ge 1$, $p_i$ prime).
Note that $C_i$ has $\phi(p_i^{a_i})=p_i^{a_i-1}(p_i-1)$ automorphisms and already by combining these, we obtain a subgroup of $\operatorname{Aut}(G)$ of order 
$$\tag2 \prod_ip_i^{a_i-1}(p_i-1).$$
As $\operatorname{Aut}(G)$ has prime order $p$, $(2)$ must be either $1$ or $p$. In particular, each  $p_i^{a_i-1}(p_i-1)$ is either $1$ or $p$. The former happens only for $a_i=1$ and $p_i=2$, the latter for $a_i=1$ and $p_i=p+1$ (so $p=2$ and $p_i=3$) or $p_i=2$ and $a_i=2$ (so $p=2$ again). Hence each $C_i$ is either $\Bbb Z/2\Bbb Z$ or $\Bbb Z/3\Bbb Z$ or $\Bbb Z/4\Bbb Z$.
So
$$G=(\Bbb Z/2\Bbb Z)^a\oplus (\Bbb Z/3\Bbb Z)^b\oplus(\Bbb Z/4\Bbb Z)^c.$$
By the results above, together with permuting equal summands, we already construct a subgroup of $\operatorname{Aut}(G)$ or order $2^b2^ca!b!c!$. This must be $1$ or $p$. So the only allowed values for $(a,b,c)$ are
$$(0,0,0), (0,0,1), (0,1,0), (1,0,0), (1,0,1), (1,1,0), (2,0,0). $$
Among the corresponding groups $G$, only the following have order $>3$:
$$\Bbb Z/4\Bbb Z, 
\quad \Bbb Z/2\Bbb Z\oplus\Bbb Z/4\Bbb Z,
\quad\Bbb Z/6\Bbb Z, 
\quad(\Bbb Z/2\Bbb Z)^2. 
$$
Of these, $\operatorname{Aut}((\Bbb Z/2\Bbb Z)^2)$ has order $6$, $\operatorname{Aut}(\Bbb Z/6\Bbb Z)$ has order $2$, $\operatorname{Aut}(\Bbb Z/2\Bbb Z\oplus\Bbb Z/4\Bbb Z)$ has order $8$, $\operatorname{Aut}(\Bbb Z/4\Bbb Z)$ has order $2$.
So we conclude that the only counter-examples to the desired claim are
$$\Bbb Z/6\Bbb Z\qquad\text{and}\qquad\Bbb Z/4\Bbb Z.$$
A: If the group order exceeds $3$, then $G$ is cyclic, in which case the automorphism group has order $\varphi(n)$, which is never prime, unless $n=4$ or $6$.  For if $G$ were not cyclic, it's automorphism group would not be abelian.  
So $G$ could be $C_4$ or $C_6$, but those are the only exceptions.
