Prove that $f$ sends transposition to transposition Let $f: S_5 \to S_5$ be authomorphism, show that then $f$ sends transposition to transposition. 
My try: 
We know every cycle is product of transposition so  for $\sigma \in S_5$ we have $f(\sigma)=f(a_1\cdot…\cdot a_k)=f(a_1)\cdot … \cdot f(a_k)$ where $a_i$ are transpositions then since $f$ authomorphism we have $o(f(a_i))|o(a_i)=2$ so $o(f(a_i)) \in\{1,2\}$ but if $o(f(a_i))=1$ for some $i$ the $f$ isn't injective so we must have f sends transp. to transp.  but I'm not sure if my reasoning is correct 
 A: Indeed, as per the comment above, an automorphism preserves the order of elements, so all you can say is that a transposition must be sent to an element of order two, so namely either another transposition or an element of $S_5$ that is written as the product of two disjoint transpositions (why are these the only elements of order 2 in $S_5$?). 
But here's one way: first, show that for any group $G$, any automorphism $\sigma$ permutes the conjugacy classes of $G$. This is just a not too hard matter of restricting an automorphism to a conjugacy class and arguing that the image is still a conjugacy class. Then, hopefully you have seen that two elements of $S_n$ are conjugate if and only if they possess the same cycle type in their disjoint cycle decomposition. Argue that the number of transpositions of $S_5$ is $5 \choose 2$ $=10$ and the number of elements in the conjugacy class of elements that are the product of two disjoint transpositions is $\frac{5}{2^22!}=15$ (you can prove these equations more generally, but here you can probably just count elements). Hence, an automorphism of $S_5$ necessarily sends the transpositions conjugacy class to itself, and you are done.
