I don't think there is a very conceptual approach giving a meaning to the comparison of conjugacy class sizes, but I think there is a way to make the argument quite short. The first thing to do is to cancel the term $a_1!$ for $i=1$ (for the points remaining fixed) in the denominator against a part of the numerator $n!$, which leaves the numerator equal to $n(n-1)\ldots(a_1+1)$ (the number of factors is the sum of the sizes of the cycles for the conjugacy class, in the group-theoretic sense where cycles cannot have length$~1$).
A first easy simplification rids us the case of mixed cycle structures, classes with is more than one different cycle length${}\geq2$. For such cycle types we can map permutations in the class to another non-trivial cycle type by simply omitting the cycle(s) of the longest length$~l$ from the decomposition. This map commutes with any conjugation by a permutations (which simple relabels elements in the cycles) and is therefore surjective to a single new conjugacy class; moreover it is not injective since there is always more than one way to reconstruct the length$~l$ cycle(s). Therefore such classes are never the smallest non-trivial ones.
The next step is to show that we need only deal with the case of transpositions, possibly flying in formation. More precisely (and with three exceptions that do not occur for $n\geq7$), the class with $m$ disjoint $k$-cycles with $k>2$ is larger than the one with $m$ disjoint $2$-cycles. We have $a_k=m$ and factor our formula as
$$
\frac{n(n-1)\ldots(n-mk+1)}{k^m}\times\frac1{m!},
$$
where the second factor is identical to the one for $k=2$. In the first factor replacing $k$ by $2$ means dropping the last $m(k-2)$ factors from the numerator and dividing the denominator by $(k/2)^m$. Using that a product of $k-2$ distinct positive integers can only be${}\leq k/2$ if $k\leq4$, we see that this decreases the first factor except when $a_1=0$, and $(m,k)\in\{(1,3),(1,4),(2,3)\}$, which happens only for $n=3,4,6$ (it explains that $3$-cycles are least numerous for $n=3$, that for $n=4$ there are as many $4$-cycles as $2$-cycles, namely $6$, and that for $n=6$ there are fewer ($40$) pairs of $3$-cycles then pairs of transpositions $(45)$).
We have reduced to the case of $m$ disjoint transpositions, which we shall compare with isolated transpositions ($m=1$). For fixed $m$, increasing $n$ leaves the denominator unchanged and increases all factors in the numerator by$~1$. This gives more relative increase for $m>1$ than for $m=1$, so if the class with $m$ disjoint transpositions is to be less numerous than that with one transposition, it must already be the case for the minimal possible value of $n$, which is $n=2m$. The formula for the number of products of $m$ disjoint transpositions when $n=2m$ is the product $1\times 3\times\cdots\times(2m-1)$ (which some people, defying the ambiguity, write as $(2m-1)!!$) which is${}\leq\binom{2m}2$ only for $m=2,3$; this explains these cases for $n=4,6$. For $n\geq7$ this does not happen any more.
$\newcommand{\Size}[1]{\lvert #1 \rvert}$Many questions in one...
You should have done some linear algebra. Then you should know that two $n \times n$ matrices $M, N$ represent the same linear map with respect to two different basis if and only if they are conjugate, that is, $g M g^{-1} = N$, where $g$ is the (invertible) matrix giving the change of bases.
As to the titular question, you should have done the orbit-stabilizer theorem, which tells you that if the finite group $G$ acts on the set $A$, and $a \in A$, then
$$
\Size{G} = \Size{a^G} \cdot \Size{G_a},
$$
where $a^G$ is the orbit and $G_a$ is the stabilizer. In your case, let $G$ act on $A = G$ by conjugation, and you are done.
Best Answer
More systematically, you have $n!$ choices to arrange $1,\ldots, n$. Place them into the parentheses pattern in this order to obtain an element of the conjugacy class. For each $r$-cycle, you divide by $r$ as only the cyclic order within a cycle plays a role, not which element we start with. Then, if there are $n_r$ cycles of length $r$, divide by $n_r!$ as the order in which the cycles are listed is not important. Note that this must also be done for the cycles of length $1$!
This gives us $$ \frac{n!}{\prod_{r}r^{n_r}n_r!} $$ Thus in $S_5$, there are $\frac{5!}{2\cdot 3}$ conjugates of $(1\,2)(3\,4\,5)$. Similarly, there are $\frac{7!}{2\cdot2\cdot 2!\cdot 3!}$ conjugates of $(1\,2)(3\,4)$ in $S_7$.