Count of Binary Operations with Fixed Identity

Theorem
Let $S$ be a set whose cardinality is $n$.

Let $x \in S$.

The number $N$ of possible different binary operations such that $x$ is an identity element that can be applied to $S$ is given by:


 * $N = n^{\left({\left({n-1}\right)^2}\right)}$

Proof
Let $S$ be a set such that $\left|{S}\right| = n$.

Let $x \in S$ be an identity element.

From Count of Binary Operations on a Set, there are $n^{\left({n^2}\right)}$ binary operations in total.

We also know that $a \in S \implies a \circ x = a = x \circ a$, so all operations on $x$ are already specified.

It remains to count all possible combinations of the remaining $n-1$ elements.

This is effectively counting the mappings $\left({S - \left\{{x}\right\}}\right) \times \left({S - \left\{{x}\right\}}\right) \to S$.

From Count of Binary Operations on a Set, this is $n^{\left({\left({n-1}\right)^2}\right)}$ structures with $x$ as the identity.

Comment
The number grows rapidly with $n$:

$\begin{array} {c|cr} n & \left({n-1}\right)^2 & n^{\left({\left({n-1}\right)^2}\right)}\\ \hline 1 & 0 & 1 \\ 2 & 1 & 2 \\ 3 & 4 & 81 \\ 4 & 9 & 262 \, 144 \\ \end{array}$