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.
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:
- 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.
- 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).
- 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).
A similar notion of pass can be used for many $a, b$ to reduce to a single case that needs to be tested. Consider the following conditions:
- $c \succeq b$
- $d \succeq b$
- $a = cq + r$ with $0 \le r < c$
- $a = ds + t$ with $0 \le t < d$
- $\tfrac{t-r}{q} \le c \bmod s \le \tfrac tq$
Then we can reduce the question of whether $a \stackrel?\succeq b$ to a single test vector as follows:
- Given an initial vector $\{u_1, \ldots, u_a\}$ we apply a pass of averages of size $c$ to reduce to $q+1$ clusters $\{v_1^c, v_2^c, \ldots, v_q^c, v_{q+1}^r\}$.
- We now apply a pass of $s$ averages of size $d$ where the input to each average is $\{v_1^{\left\lfloor c/s \right\rfloor}, v_2^{\left\lfloor c/s \right\rfloor}, \ldots, v_q^{\left\lfloor c/s \right\rfloor}, v_{q+1}^{d - q\left\lfloor c/s \right\rfloor} \}$
- That second pass creates one value $w_1$ with frequency $ds$. A final average of size $d$ which includes the $t$ values not equal to $w_1$ (and $w_1^{d-t}$) reduces to $\{w_1^{a-d}, w_2^d\}$, which by linearity is equivalent to $\{0^{a-d}, 1^d\}$. So if that latter test vector can be averaged, $a \succeq b$. (And, clearly, if it can't then $a \not\succeq b$).
The validity of step 2 depends on $q\left\lfloor \frac cs \right\rfloor \le d$; since $s\left\lfloor \frac cs \right\rfloor = c - (c \bmod s)$ this is equivalent to $$\begin{eqnarray*} cq - q(c \bmod s) &\le& ds \\ a - r - q(c \bmod s) &\le& a - t \\ t - r &\le& q(c \bmod s) \\ \end{eqnarray*}$$
We also require there to be sufficient $v_{q+1}$; i.e. $$\begin{eqnarray*} s(d - q\left\lfloor \frac cs \right\rfloor) &\le& r \\ ds - cq + q (c \bmod s) &\le& r \\ q (c \bmod s) &\le& t \\ \end{eqnarray*}$$
Combining the two, we see the necessity for the final precondition.
I've done a bit of experimentation with solving the test vectors produced by this reduction. If we consider the graph whose vertices are multisets of $a$ rationals and whose edges correspond to averaging $b$ values, solving a test vector is a pathfinding problem. However, the graph is infinite and most vertices have quite large out-degrees. The vertex priority which I've found most effective is to use Ilya Bogdanov's technique for normalising to a multiset of integers, and then prioritise the multisets with smallest maximum absolute value. Updated code and solutions will appear shortly on the gist I posted earlier in comments.