The 'simpler', the weirder

Let us first consider this problem with a finite set $S$ instead of $\mathbb N$ , with n( $S$ )=n

WLOG, we take $f$ as a reference function.

Let $f=({(s_1,s'_1),(s_2,s'_2),....,(s_n,s'_n)})$ , with ${s_i}\in S$ and ${s'_i}$ is the unique image of ${s_i}$ under $f$ (since $f$ is invertible). Now, for $g$ to intersect $f$ at exactly $k$ points, any $k$ of the $({s_i,s'_i}$ ) pairs in $f$ should be same as in $g$ (which is also invertible).

Number of ways of choosing these = $n\choose k$

Now, rest of the ( $n-k$ ) items need to be deranged, as none of the other pairs should match.

Derangement of $n-k$ items = $(n-k)!D(n-k)$ where $D(n-k)$ = $\sum_{r=0}^{n-k}\frac{(-1)^r}{r!}$

$\implies$ number of functions $g$ possible under given conditions = ${n\choose k} [(n-k)!D(n-k)]$

Total number of ways of forming $g = n!$

$\implies$ Probability $(P)$ of $g$ intersecting $f$ at $k$ points = $\frac{{n\choose k} (n-k)!D(n-k)}{n!} = \frac{D(n-k)}{k!}$

Now, replacing $S$ with $\mathbb N$ (for our case) and since $n(\mathbb N)→∞$ , $P→\frac{1}{k! e}$
(using Taylor series expansion $↣D(n-k) →\frac{1}{e}$ as $n→∞$ )

$\implies A (answer) = \sum_{i=0}^{k} P.$ Substituting $k=4$ and $P$ , $A = \frac{1}{e}(\frac{1}{0!}+\frac{1}{1!}+..+\frac{1}{4!}) = 0.9963401..$ $\implies \lfloor 10^5A\rfloor = \boxed{99634}$

Will refine any ambiguities soon :).