Show that not all 5 cycles in $A_5$ are conjugate in $A_5$

Let me suggest a correction on the approach you're talking about in the comments:

• Start with a $5$-cycle $(a\:b\:c\:d\:e).$
• Show that there exists another $5$-cycle $(f\:g\:h\:i\:j)$ such that for all $r\in A_5,$ we have $$\bigl(r(a)\:r(b)\:r(c)\:r(d)\:r(e)\bigr)\ne (f\:g\:h\:i\:j)$$

This is a rather cumbersome approach, though, especially since there are $5$ equivalent but distinct ways to represent a given $5$-cycle with this notation. We could also guess and check, but that is even more cumbersome.

If you know that the order of a conjugacy class must divide the order of the group, then we can proceed much more simply. How many $5$-cycles are there? What is the order of $A_5$?

The conjugacy map on a 5-cycle $(a,b,c,d,e)$ (or any cycle) by $\sigma$ is obtained on replacing each element of the cycle by $\sigma$ applied to that element. So select some $\sigma$ in $S_5$ but not in $A_5$, for example the 2-cycle $(a,b)$, and conjugate a given 5-cycle by that.

Specific example: The two cycles $(12345)$ and $(21345)$ cannot be conjugate in $A_5$. (The second cycle can of course be rewritten as $(13452)$ if you want it to start with the lowest entry as is usually done.)

ADDED: as SteveD has noted in a comment, this needs a further check. The cycle (21345) may be written in 5 ways, each of which has to be checked to see whether the conjugating map from 12345 to it is in $A_5$. It turns out non e of them are:

$$12345 \to 21345 : (12)\\ 12345 \to 13452 : (2345) \\ 12345 \to 34521: (135)(24) \\ 12345 \to 45213 : (14)(253) \\ 12345 \to 52134 : (1543).$$ In each case the cycle form of the conjugating map lies in $S_5-A_5$, i.e. the conjugating map is an odd permutation. This means there is no conjugation of $(12345)$ to $(21345)$ by an element of $A_5.$

Here is a quite general lemma, you might be interested in.

Let $G$ be a finite group acting transitively on a set $\Omega$ and let $N \trianglelefteq G$ be a normal subgroup. Then the number of $N$-orbits of $\Omega$ equals $|G : N G_\omega|$ where $G_\omega$ is the stabilizer of an arbitrary $\omega \in \Omega$ in $G$. Furthermore they all have the same length.

Consider now $G = S_n$ and let $\Omega$ be the set of all permutations of $S_n$ with the same given cycle type (e.g. all cycles of length 5). It is well known that $S_n$ acts transitively on $\Omega$ by conjugation, so the above lemma tells us that $\Omega$ splits into $|S_n : A_n C_\sigma| \in \{1,2\}$ $A_n$-orbits where $C_\sigma$ is the centralizer of some permutation $\sigma \in \Omega$. In particular there are elements in $\Omega$ which are not conjugated in $A_n$ if and only if the centralizer of one (and hence all) permutation of $\Omega$ is contained in $A_n$.

From the proof of that lemma we see that in this case $\sigma$ and $\sigma^\tau$ are not conjugated in $A_n$ for any $\sigma \in \Omega$ and $\tau \in S_n \setminus A_n$.