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 $$n$$:

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