Exotic map from $S_5$ to $S_6$

a) Show that if $\phi$ from $S_5$ to $S_6$ is any exotic map, then $\phi$ is injective, and moreover $phi$ doesn't send transpositions to transpositions.

b) If $\phi$ is as in part a), show that $\phi(A_5)$ is contained in $A_6$ and conclude that $H$=$Im(\phi)$ contains no transpositions.

Admittedly I couldn't make much progress in this problem, primarily because I don't really understand the exotic map to start with. I scoured around the internet but information on exotic map is really scarce and not very useful.

I also looked at the example of the mystic pentagons, but it wasn't of much help really.

Any help much appreciated. Especially if someone can explain exotic map to me that'd be so great.

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Isn't there some more notes around (before and after) the exersize? Exotic should be in contrast to some kind of easily arosen homomorphisms $S_5\to S_6$. –  Berci Nov 11 '12 at 23:53
There is, but all it says is "A group homomorphism with the property that the natural action of $S_5$ on {1,2,3,4,5,6} induced by $\phi$ is transitive. –  Benjamin Lu Nov 12 '12 at 0:09
You will find a lot of what you want by searching for 'outer automorphism' and looking for the results whose titles contain $S_6$. math.stackexchange.com/search?q=outer+automorphism –  Alexander Gruber Nov 12 '12 at 6:15

I assume that 'exotic map' means what you wrote in the comment: $\phi:S_5\to S_6$ acts transitively on $\{1,..,6\}$, that is: for each pair $(a,b)$ there is a $g\in S_5$ such that $\left(\phi(g)\right)(a)=b$.
1. Assume that $\phi$ maps a transposition $(a,b)$ to transposition, then, as any transposition is a conjugate of $(a,b)$ in $S_5$, we have that in fact $\phi$ then must map every transposition to a transposition. Now, if transpositions $(a,b)$ and $(b,c)$ intersect in $S_5$, then $(a,b)(b,c)=(a,b,c)$ has order $3$, and if their $\phi$-images were disjoint transpositions, their product would have order $2$. So that, the following transpositions in $S_6$ intersect: $\phi(1,2),\ \phi(1,3),\ \phi(1,4), \phi(1,5)$. It means that they may move maximum $5$ elements among the six, but these transpositions already generate $S_5$, thus $\phi$ is not transitive.
2. To show that $\phi$ is injective, consider its kernel: obviously it cannot be $S_5$ (because then $\phi$ could not be transitive at all), but neither $A_5$, because then $im\phi\cong S_5/A_5\cong \Bbb Z_2$ would be, so again, the action is not transitive. And there are no more normal subgroups of $S_5$.
3. Similarly, as the preimage of a normalsubgroup, $\phi^{-1}(A_6)$ is normal in $S_5$, if it is either $A_5$ or $S_5$, then $\phi(A_5)\subseteq A_6$ as wished. The remaining case is when $\phi^{-1}(A_6)=\{id\}$. This is going to contradict again, either to 2. or to the transitivity.
4. Suppose that $\phi(g)=(a,b)$ transposition for some $g\in S_5$. As $\phi(g^2)=id$, and $\phi$ is injective, we have $o(g)=2$, that is, the cycle structure of $g$ must be some disjoint transposition. But, in $S_5$ we can have maximum $2$ disjoint transpositions(!), so either $g=(u,v)$ or $g=(u,v)(w,z)$. But the latter one is in $A_5$, while $\phi(g)\notin A_6$, and the first one was excluded at 1.