Question: How can I get the cardinality of symmetric group $S_X$?
I have found similar posts but they do not seam to match for me because there the advice is given to calculate the cardinality like $k^k$.
The symmetric group I am talking about is defined as:
$S_X:= Map(X, X)^X$
Example:
$S_0=\{id_{\emptyset}\}=\{()\}$
$S_1=\{id_1\}= \{ \left( \begin{array}{cc|c} 1\\ 1 \end{array} \right) \}$
$S_2=\{id_{1,2}\}= \{ \left( \begin{array}{cc|c} 1 & 2\\ 2 & 1 \end{array} \right) , \left( \begin{array}{cc|c} 2 & 1\\ 1 & 2 \end{array} \right) \}$
$S_3=\{id_{1,2,3}\}= \{ \left( \begin{array}{ccc} 1 & 2 & 3\\ 1 & 2 & 3 \end{array} \right) , \left( \begin{array}{ccc} 1 & 2 & 3\\ 2 & 1 & 3 \end{array} \right), \left( \begin{array}{ccc} 1 & 2 & 3\\ 3 & 2 & 1 \end{array} \right), \left( \begin{array}{ccc} 1 & 2 & 3\\ 1 & 3 & 2 \end{array} \right), \left( \begin{array}{ccc} 1 & 2 & 3\\ 2 & 3 & 1 \end{array} \right), \left( \begin{array}{ccc} 1 & 2 & 3\\ 3 & 1 & 1 \end{array} \right) \}$
The second row in any element of $S_n$ is a permutation of the set $\{1, 2, 3, \dots, n\}$, of which there are $n!$.
One way you could go about finding this out yourself is to compute the number of elements in $S_n$ for small $n$ (e.g. 1, 2, 3, ... which you have done already) and then putting them into the OEIS.
The reason the answer of $k^k$ does not work here is because the answer you linked is for infinite sets, where $n!$ (or in this case $k!$) would involve an infinite product, where $k$ of the factors are of size $k$ because $k$ is infinite (This is not a rigorous demonstration of $k^k$, but it is designed to give some intuition on why this would be the case if you are familiar with the notion of cardinality.).