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.
There are at least two possible proofs, one of them by enumeration and
another one using the exponential formula.
By enumeration, first choose the elements to go on each cycle
(multinomial coefficient):
$$\frac{N!}{\prod_j (j!)^{P_j}}$$
Each of these elements generates $j!/j$ cycles (in placing labels on a
directed cycle all orbits under the action of the cyclic group have
the same size which is $j$):
$$\prod_j \frac{(j!)^{P_j}}{j^{P_j}}$$
Permutations of the size $j$ cycles are not distinguished:
$$\prod_j \frac{1}{P_j!}.$$
This yields the answer
$$\frac{N!}{\prod_j (j!)^{P_j}}
\prod_j \frac{(j!)^{P_j}}{j^{P_j}}
\prod_j \frac{1}{P_j!}
= \frac{N!}{\prod_j j^{P_j} P_j!}.$$
The second proof uses the exponential formula (OGF of the cycle index
of the symmetric group)
$$Z(S_N) = [z^N]
\exp\left(a_1 z + a_2 \frac{z^2}{2} + a_3 \frac{z^3}{3} + \cdots\right).$$
Extracting the coefficient of $[z^N a_1^{P_1} a_2^{P_2} \times\cdots]$
in $N! Z(S_N)$ we get
$$N! [z^N] [\prod_j a_j^{P_j}] \prod_j \exp\left(a_j\frac{z^j}{j}\right)
= N! [z^N] \prod_j \frac{z^{j P_j}}{j^{P_j} P_j!}
\\ = N! [z^N] z^{\sum_j jP_j} \prod_j \frac{1}{j^{P_j} P_j!}
= N! \prod_{j} \frac{1}{j^{P_j} P_j!},$$
the same as in the first version. (Here we take $P_j = 0$ for a cycle
size that is absent.)
Remark. The reason for treating the OGF like an EGF is that we
have the labeled species
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}(
\mathcal{A_1}\textsc{CYC}_{=1}(\mathcal{Z})
+ \mathcal{A_2}\textsc{CYC}_{=2}(\mathcal{Z})
+ \mathcal{A_3}\textsc{CYC}_{=3}(\mathcal{Z})
+ \cdots)$$
and hence we are extracting from an EGF.
Best Answer
Let's create some group as an example:
It has 8 conjugacy classes of elements:
Let's take the 2nd of them:
It contains 180 elements:
and you can test membership of group elements in the conjugacy class as follows:
Conjugacy class is not internally represented as a list of its elements - that would be very inefficient (for example, for algorithms that need only representatives of conjugacy classes). But if you need to get a list if all elements of the class, you can get them as follows:
This may be very memory inefficient for large groups, but you can also iterate over its elements as follows:
without constructing the whole list, and also use enumerator which will give you a list-like behaviour:
also without constructing the whole list.