Let $p > 2$ be a prime number. $\mathbb{F}_p[x]$ is defined as the set of polynomials in $x$ with coefficients in $\mathbb{F}_p$ (the integers modulo $p$ with usual addition and subtraction), so that two polynomials are equal if and only if the coefficients of $x^k$ are equal in $\mathbb{F}_p$ for each nonnegative integer $k$. For example, $(x+2)(2x+3) = 2x^2 + 2x + 1$ in $\mathbb{F}_5[x]$ because the corresponding coefficients are equal modulo 5.

Let $f, g \in \mathbb{F}_p[x]$. The pair $(f, g)$ is called [i]compositional[/i] if
\[f(g(x)) \equiv x^{p^2} - x\]
in $\mathbb{F}_p[x]$. Find, with proof, the number of compositional pairs.
To find the number of compositional pairs \((f, g)\) such that \(f(g(x)) \equiv x^{p^2} - x\) in \(\mathbb{F}_p[x]\), we will consider the degrees of \(f\) and \(g\). Clearly, \(\left(\deg f, \deg g\right)\) must be one of the following:
\[
\left\{(1, p^2), (p^2, 1), (p, p)\right\}
\]

### Case 1: \(\deg f = 1\)
Let \(f(x) = ax + b\) with \(a \neq 0\). Then we have:
\[
f(g(x)) = a g(x) + b = x^{p^2} - x
\]
This implies:
\[
a g(x) + b = x^{p^2} - x
\]
Solving for \(g(x)\), we get:
\[
g(x) = \frac{x^{p^2} - x - b}{a}
\]
Since \(a \neq 0\) and \(b\) can be any element in \(\mathbb{F}_p\), there are \(p(p-1)\) choices for \(a\) and \(b\).

### Case 2: \(\deg g = 1\)
Let \(g(x) = ax + b\) with \(a \neq 0\). Then we have:
\[
f(ax + b) = x^{p^2} - x
\]
Letting \(y = ax + b\), we get:
\[
f(y) = \left(\frac{y - b}{a}\right)^{p^2} - \frac{y - b}{a} = \frac{1}{a} \left((y - b)^{p^2} - (y - b)\right)
\]
Since \(a \neq 0\) and \(b\) can be any element in \(\mathbb{F}_p\), there are \(p(p-1)\) choices for \(a\) and \(b\).

### Case 3: \(\deg f = \deg g = p\)
We take the derivative of \(f \circ g\) with respect to \(x\):
\[
f'(g(x)) g'(x) = -1
\]
Since \(\mathbb{F}_p\) is a field, we must have \(g'(x) = a\) for some non-zero constant \(a\). Then:
\[
f'(g(x)) = -\frac{1}{a}
\]
A polynomial in \(t\) has zero derivative in \(\mathbb{F}_p\) if and only if its exponents are divisible by \(p\). Thus, the exponents of \(g(x) - ax\) are divisible by \(p\). Since \(\deg g = p\), we have:
\[
g(x) = bx^p + ax + c
\]
for some constants \(b \neq 0\) and \(c\). Similarly, the exponents of \(f(g(x)) + \frac{1}{a}x\) (as a polynomial in \(g(x)\)) are divisible by \(p\). Since \(\deg f = p\), we have:
\[
f(g(x)) = dg(x)^p - \frac{1}{a}g(x) + e = d(g(x)^p) - \frac{1}{a}g(x) + e
\]
for some constants \(d \neq 0\) and \(e\). Thus, we have:
\[
x^{p^2} - x = f(g(x)) = d(bx^{p^2} + ax^p + c) - \frac{1}{a}(bx^p + ax + c) + e
\]
Simplifying, we get:
\[
x^{p^2} - x = dbx^{p^2} + (ad - \frac{b}{a})x^p - x + (cd - \frac{c}{a} + e)
\]
Matching coefficients, we get:
\[
db = 1, \quad ad = \frac{b}{a}, \quad e = \frac{c}{a} - cd
\]
Choosing \(b \neq 0\) and \(c\) arbitrarily, we have \(a = \pm b\), \(d = \frac{1}{b}\), and \(e = \frac{c}{a} - cd\). Thus, there are \(2p(p-1)\) choices here.

Combining these cases, we deduce that there are:
\[
\boxed{4p(p-1)}
\]
choices of \((f, g)\).