Let G be a group of order 12. Does the group of automorphisms of G contain an element of order 5? 
Let $G$ be a group of order $12$. Does $\mathrm{Aut}(G)$ contain an element of order $5$?  

I have tried using Sylow Theorems to obtain (G has only one Sylow 2-subgroup (order 4) or G has only one Sylow 3-subgroup (order 3)), G is not simple. And I was trying to construct some group actions, but these seem to lead me to nowhere.  
 A: Here is an argument that works for all groups of order $12$.
Suppose that $\alpha \in {\rm Aut}(G)$ of order $5$ with $|G|=12$. Since the elements in $G$ that are fixed by $\alpha$ form a subgroup, $\alpha$ must act on the  elements of $G$ with two orbits of length $5$ and fixed subgroup of order $2$.
The elements in the same orbit have the same order, and there exist elements of order $3$. The total number of elements of order $3$ is even, so there must be $10$ such, but that would give $5$ Sylow $3$-subgroups, contradicting Sylow's theorem. 
A: Take for example the cyclic group of order $\;12\;$ :
$$\text{Aut}\,(C_{12})\cong C_{12}^*\cong C_3^*\times C_4^*\cong C_2\times C_2$$
and thus we don't even have an element of order five.
Or even simpler using another approach: it is always true that $\;|\text{Aut}\,(C_n)|=\phi(n)\;,\;\;\phi=$ Euler's Totien Functions, so for the cyclic group of order $\;12\;$ we have $\;\phi(12)=\phi(3)\phi(4)=2\cdot2=4\;$ , and a group of order four cannot have an element of order five (why?).
A: With the hints and guidance from some helpful teaching assistants, I came up with the following: 
Let H be the cyclic subgroup generated by an automorphism on G of order 5, let X be the set of all Sylow 2-subgroups of G, Y be the set of all Sylow 3-subgroups of G.  A group action of H on X is trivial.  So another action of H on a fixed point (that is a Sylow 2-subgroup, denote it by $P_2$) of X could be defined and the second action is also trivial.  So H fixed at least four elements of order 2 in G.  For the remaining elements in G that belong to a Sylow 3-subgroup, $P_3$ of G, the elements are as well fixed by H. Since any automorphism in H is a homomorphism, H also fixes elements in $P_2 \cdot P_3$. Thus any automorphism in H actually fixes all elements in G, and is thus an identity map.  An identity map in Aut(G) is of order one, not of order five.  Contradiction arises. 
