The statement you make about tetrahedra generalizes to arbitrary $n$. Specifically, the symmetric group $S_n$ is the group of symmetries of a regular $n$-simplex, and the alternating group $A_n$ acts on this simplex by rotations. (In fact, $A_n$ is precisely the set of rotational symmetries.) Elements of different conjugacy classes of $A_n$ are geometrically distinguishable, in the sense that they would "look different" to an observer in $\mathbb{R}^n$.
One way of making the notion of "look different" precise is that non-conjugate elements of $A_n$ correspond to non-conjugate elements of the rotation group $SO(n)$. Thus, two non-conjugate elements of $A_n$ do not look the same up to rotation of the simplex.
Incidentally, the simplest algorithm to distinguish conjugacy classes in $A_n$ is essentially to check the sign of the conjugator. For example, the elements $(5)(2\;6\;3)(1\;9\;4\;8\;7)$ and $(2)(1\;4\;8)(3\;7\;5\;6\;9)$ are conjugate in $S_9$, and their conjugacy class in $S_9$ splits into two conjugacy classes in $A_9$. To check whether the two elements are conjugate in $A_9$, we consider a permutation that maps between corresponding numbers:
$$
\begin{bmatrix}
5 & 2 & 6 & 3 & 1 & 9 & 4 & 8 & 7\\
2 & 1 & 4 & 8 & 3 & 7 & 5 & 6 & 9
\end{bmatrix} \;=\; (1\;3\;8\;6\;4\;5\;2)(7\;9)
$$
This permutation is odd, so the two elements are not in the same conjugacy class in $A_9$. It is possible to construe this algorithm geometrically, since the difference between odd and even permutations is the same as the difference between right-handed and left-handed coordinate systems.
Consider the effect of conjugation on cycles. Given the cycle $c=(a_1~a_2~\cdots~a_k)$ and $\sigma\in S_n$, the permutation $\sigma c\sigma^{-1}$ maps $\sigma(a_i)$ to $\sigma(a_{i+1})$ (cycle the index modulo $k$), and all other arguments are fixed by the permutation. Hence
$$c^\sigma=\sigma(a_1~a_2~\cdots~a_k)\sigma^{-1}=(\sigma (a_1)~\sigma(a_2)~\cdots~\sigma(a_k)).$$
If $c_1,\cdots,c_l$ are disjoint cycles, $\sigma(c_1\cdots c_l)\sigma^{-1}=(\sigma c_1\sigma^{-1})\cdots(\sigma c_l\sigma^{-1})$, hence conjugation preserves not only cycles and cycle lengths, but cycle types as well. Moreover, conjugation acts transitively on permutations of a given cycle type. Suppose $\pi=c_1\cdots c_l$ and $\rho=d_1\cdots d_l$ are two permutations and their disjoint cycle decompositions such that each $c_i$ and $d_i$ have the same length.
We construct a $\sigma$ in pieces: if $c_i=(a_1~a_2~\cdots~a_k)$ and $d_i=(b_1~b_2~\cdots~b_k)$ (note $k$ can vary with $i$), let $\sigma$ send $a_j\mapsto b_j$ for $j=1,2,\cdots,k$. After doing this for each cycle (even the ones of length one, which are normally left out of written representations for efficiency), we will have that $\sigma \pi \sigma^{-1}=\rho$, by having equivalent cycle decompositions. This $\sigma$ is not unique; we could rewrite $c_i$ cycled by $1$ in index as $(a_k~a_1~\cdots~a_{k-1})$ and $\sigma$ would turn out different.
Cycle types are in bijection with integer partitions. For $n=5$, we have
$$\begin{array}{c | c}\lambda\vdash5 & \sigma \\ \hline (5) & (1~2~3~4~5) \\
(4,1) & (1~2~3~4)(5) \\
(3,2) & (1~2~3)(4~5) \\
(3,1,1) & (1~2~3)(4)(5) \\
(2,2,1) & (1~2)(3~4)(5) \\
(2,1,1,1) & (1~2)(3)(4)(5) \\
(1,1,1,1,1) & (1)(2)(3)(4)(5) \end{array}$$
On the left are integer partitions of $5$, and on the right are easily constructed representatives (in cycle notation) of the conjugacy class - cycle type - associated to the partition.
Best Answer
The conjugacy classes of $A_5$ are the orbits of the action of $A_5$ in $A_5$ given by the conjugacy action. You know that in $S_5$ everyone of those elements are in an unique conjugacy class and they represent all the classes. You know from the class equation that (in $S_5$)$$|\mathrm{orb}(x)_{S_5}|=\frac{|S_5|}{|\mathrm{Stab}(x)_{S_5}|}$$
We are trying to study $\mathrm{orb}(x)_{A_5}$.
Since $\mathrm{Stab}(x)_{S_5} < S_5$, we have two possibilities:
1) $\mathrm{Stab}(x)_{S_5} \subseteq A_5$: in this case $|\mathrm{orb}(x)_{A_5}|=\frac{1}{2}|\mathrm{orb}(x)_{S_5}|$, so your class in $S_5$ splits in two new classes in $A_5$.
2) $\mathrm{Stab}(x)_{S_5} \nsubseteq A_5$: since $A_5$ is a subgroup of index 2, and since $$\mathrm{Stab}(x)_{A_5}=A_5 \cap \mathrm{Stab}(x)_{S_5}$$ you get$$[\mathrm{Stab}(x)_{A_5}:\mathrm{Stab}(x)_{S_5}]=2$$ So you have $|\mathrm{orb}(x)_{A_5}|=|\mathrm{orb}(x)_{S_5}|$, and you get the same conjugacy class.
Moral: you just have to know if $\exists \tau \in \mathrm{Stab}(x)_{S_5} | \tau \notin A_5$ i.e.: if we are interested in studying if the conjugacy class of $(123)$ splits when you go in $A_5$, bacause $(45) \in \mathrm{Stab}((123))_{S_5} \smallsetminus A_5 $ you know that it DOES NOT SPLIT.
Was I clear?