, 2024

Solution:
Let $f g$ represent the composition of permutations $f$ and $g$, where $(f g)(x)=f(g(x))$ for all $x \in \{1,2,3,4,5\}$.

Evaluating $f g h f h$ in two ways, we get
$$f = g f h = (f g h) f h = f g h f h = f(g h f) h = f h h,$$
so $h h = 1$. Similarly, we get $f, g$, and $h$ are all involutions. Then
$$f g h = g \Longrightarrow f g = g h,$$
so $f g = g h = h f$. Let $x := f g = g h = h f$. Then
$$x^{3} = (f g)(g h)(h f) = 1$$
We can also show that $f g = g h = h f$ along with $f, g, h$ being involutions is enough to recover the initial conditions, so we focus on satisfying these new conditions.

If $x = 1$, then $f = g = h$ is an involution. There are $1 + \binom{5}{2} + \frac{1}{2} \binom{5}{2,2,1} = 26$ involutions, so this case gives 26 solutions.

Suppose $x \neq 1$. Then since $x^{3} = 1$, $x$ is composed of a 3-cycle and two fixed points, of which there are 20 choices. WLOG $x = (123)$. It can be checked that $\{1,2,3\}$ must map to itself for all of $f, g, h$ and also $\{4,5\}$. We can either have all of $f, g, h$ map 4 and 5 to themselves or each other. Restricted to $\{1,2,3\}$, they are some rotation of (12), (23), (13). Each of the 20 cases thus gives $2 \cdot 3 = 6$ triples, so overall we get $20 \cdot 6 = 120$.

The final answer is $26 + 120 = 146$.