Q: Why is that some polynomial relations between eta quotients have a solvable Galois group, even if the deg is $n>4$?
For example, we have the well-known modular equation,
$$u^6 - v^6 + 5u^2v^2(u^2 - v^2) + 4uv(1 - u^4v^4)=0\tag{1}$$
which is satisfied by,
$$u = \frac{\sqrt{2}\,\eta(\tau)\eta^2(4\tau)}{\eta^3(2\tau)},\;\;v = \frac{\sqrt{2}\,\eta(5\tau)\eta^2(20\tau)}{\eta^3(10\tau)}$$
but does not have a solvable Galois group. In contrast,
$(x + 7 x^2 + 21 x^3 + 49 x^4 + 147 x^5 + 343 x^6 + 343 x^7) \\+ 7(x + 5 x^2 + 7 x^3) m- m^2 = 0\tag{2}$
which is the relation between,
$$x = \frac{\eta(49\tau)}{\eta(\tau)},\;\;m = \left(\frac{\eta(7\tau)}{\eta(\tau)}\right)^4$$
as a 7th deg eqn in $x$ has a solvable Galois group for any constant $m$. It has order 42, discriminant $D=-7^{17} (1+5m+m^2)^2(1+13m+49m^2)^4$, hence is one of the few known examples of parametric solvable septics. Let $m=1$, then,
$$-1 + 8 x + 42 x^2 + 70 x^3 + 49 x^4 + 147 x^5 + 343 x^6 + 343 x^7=0$$
and so on for all $m$. I found one can do so for $p = 3,5,7,13$. For $p=13$ and any constant $m$, define,
$$f_n = \frac{x(x^n - 1)}{x - 1}m^{p - n}$$
then,
$$f_{13} + 13 (2 f_{12} + 25 f_{11} + 196 f_{10} + 1064 f_9 + 4180 f_8 + 12086 f_7 + 25660 f_6 + 39182 f_5 + 41140 f_4 + 27272 f_3 + 9604 f_2 + 1165 f_1) - 13m^{14} = 0\tag{3} $$
As an equation in $x$ of deg $13$, this has a solvable Galois group of order $12\cdot13 = 156$. For example, let $m=1$, then,
$$-13 + 2100489 x + 2085344 x^2 + 1960492 x^3 + 1605956 x^4 + 1071136 x^5 + 561770 x^6 + 228190 x^7 + 71072 x^8 + 16732 x^9 + 2900 x^{10} + 352 x^{11} + 27 x^{12} + x^{13}=0$$
is solvable in radicals (as is for all $m$). But (3) is also the relationship between,
$$x = \left(\frac{\sqrt{13}\eta(169\tau)}{\eta(\tau)}\right)^2,\;\; m = \left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^2$$
Q: Would this imply analogous eta quotients $\frac{\eta(p\tau)}{\eta(\tau)}$ for $p>13$ would also involve equations with a solvable Galois group?
