A brief comment on notation: "can be averaged by" is a reflexive partial order. I shall use the notation $a \succeq b$ to indicate that $a$ can be averaged by $b$, and save $\geq$ for the usual total order on the integers.

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Another partial answer, this time with a positive proposition which I'm calling the **three-pass lemma**:

> If $a = mn$, $\exists s < n: sm \succeq b$, and $\exists t < m: tn \succeq b$ then $a \succeq b$.

Proof proceeds by arranging the $a$ values in $m$ rows of $n$ values each and then performing three passes:

1. Average the first $t$ rows, then the next $t$ rows, etc. If $tn \not\mid a$, finish by averaging the last $t$ rows. At the end of this pass, every row contains a single value repeated $n$ times.
2. Average the first $s$ columns, then the next $s$ columns, etc. If $sm \not\mid a$, finish by averaging the last $s$ columns. At the end of this pass, there only two distinct values remain: the first $n-s$ columns have one value, and the last $s$ columns have another. (If $sm \mid a$ then only one distinct value remains and we can stop).
3. Repeat the first pass. Each averaging operation of this pass takes the averaged positions to the target value.

Note that this proposition generalises your property 2 (take $m=p$, $n=a$, $s = \tfrac bp$, $t=1$) as well as your property 4 (take $m=n=y$, $s=t=x$).

**Corollary**:

> Given $a$, we can find $b < a$ such that $b \preceq a$ iff $a$ is not squarefree.

Clearly if $a$ is squarefree then from $p \mid a \implies p \mid b$ the smallest predecessor of $a$ is itself. On the other hand, if $a = p^2 a'$ then we can take $m = p$, $n = pa'$, $s = a'$, $t = 1$, $b = pa'$ in the three-pass lemma. (This is also an instance of your property 2).