Given $\large q=e^{2\pi i \tau}$. Define,
$$\alpha(\tau) = \sqrt2\,q^{1/8}\prod\_{n=1}^\infty\frac{ (1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}$$
$$\beta(\tau) = q^{1/5}\prod\_{n=1}^\infty\frac{ (1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$
$$\gamma(\tau) = q^{1/4}\prod\_{n=1}^\infty\frac{ (1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-2})(1-q^{6n-4})}$$
$$\delta(\tau) = q^{1/3}\prod\_{n=1}^\infty\frac{ (1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}$$
$$\epsilon(\tau) = q^{1/2}\prod\_{n=1}^\infty\frac{ (1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}$$
$$\lambda(\tau) = q^{1/1}\prod\_{n=1}^\infty\frac{ (1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}$$


each of which has a beautiful [$q$-continued fraction](https://mathoverflow.net/questions/108926/the-complete-list-of-continued-fractions-like-the-rogers-ramanujan). Then we propose the *[functional equations](https://en.wikipedia.org/wiki/Functional_equation)*,


$$\begin{aligned}
\alpha^8\Big(\frac{-1}{4\tau}\Big)&=1-\alpha^8(\tau)\\[2mm]
\beta^5\Big(\frac{-1}{5\tau}\Big)&=\frac{\phi^5\beta^5(\tau)-1}{-\beta^5(\tau)-\phi^5},\quad\phi=\tfrac{1+\sqrt5}2\\[2mm]
\gamma^4\Big(\frac{-1}{6\tau}\Big)&=\frac{\gamma^4(\tau)-\tfrac19}{\gamma^4(\tau)-1}\\[2mm]
\delta^3\Big(\frac{-1}{6\tau}\Big)&=\frac{\delta^3(\tau)-\tfrac18}{-\delta^3(\tau)-1}\\[2mm]
\epsilon^2\Big(\frac{-1}{8\tau}\Big)&=\frac{u^2\,\epsilon^2(\tau)-1}{\epsilon^2(\tau)-u^2},\quad u =1+\sqrt2\\[2mm]
\lambda\Big(\frac{-1}{12\tau}\Big)&=\frac{v\,\lambda(\tau)-1}{\lambda(\tau)-v},\quad v = 2+\sqrt3\end{aligned}$$


*Note 1*: Excepting $\gamma^4\big(\tfrac{-1}{6\tau}\big)$ which is [in this post](https://math.stackexchange.com/questions/1790253/modular-equation-for-a-quotient-of-eta-functions) and served as the model, I just found the rest empirically, but their consistent forms suggest these are correct.


*Note 2*: Also, the two order $6$ obey the simple $\displaystyle \frac1{\gamma^4(\tau)}-\frac1{\delta^3(\tau)}=1$.


**Questions:**


1. How do we derive the easiest one from first principles and generalize it for the rest?
2. Is the list complete? Without requiring representation as a q-continued fraction, is there a similar functional equation for, say, a $p=7$ order?



Some nice functional equations for $q$-continued fractions
Throughout this answer, I'll be referencing Kubert-Lang, "Modular Units" Chapter 2, sections 1 and 2, and Chapter 3 section 4.


Each of your functions is a ratio of the Siegel functions of the form 
$$A\_N(\tau)=\left(\frac{g\_{\frac1N,0}}{g\_{\frac aN,0}}(N\tau)\right)^m.$$


If $r=(r\_1,r\_2)\in \mathbb Q^2$, $B\_2(x)=x^2-x+1/6$ is the second Bernoulli polynomial, $q=e^{2\pi i \tau}$ as usual, and $\zeta=e^{2\pi i r\_2}$, then the Siegel function $g\_a$ is defined by
$$
g\_a(\tau)= -q^{\frac12 B\_2(r\_1)}e^{2\pi i \frac{r\_2(r\_1-1)}2}\prod\_{n=0}^\infty(1-q^{n+r\_1}\zeta)(1-q^{n+1-r\_1}\zeta^{-1}).
$$
The Siegel functions $g\_a$ are modular units, meaning they're modular functions whose zeros and poles are supported at cusps. 


If $\gamma\in \text{SL}\_2(\mathbb Z)$, then 
$$
g\_a|\_0\gamma=g\_{a\gamma}.
$$
Moreover, we can reduce $r\_1$ and $r\_2$ modulo $\mathbb Z$, or change signs of both $r\_1$ and $r\_2$, if we introduce a root of unity. However, the $m$'s in your ratios satisfy $m(1-a^2)\equiv 0 \pmod N$ if $N$ is odd and $\pmod{2N}$ if $N$, and so using Theorem 3.4.1 of Kubert-Lang we find we can ignore this extra factor so the transformations rules becomes this simple. Note, the theorem is written in terms of Klein functions $t\_a=g\_a\Delta^{-\frac1{12}}$. Using these transformation rules, it's easy to see that all of your functions are invariant under $\Gamma\_1(N)/\{\pm1\}$. 


The specific groups you have considered are ones where this group is genus $0$, and your functions $A\_N$ are Hauptmoduln with a unique zero at the cusp $\infty$, a constant $C\_0$ at the cusp $0$, and a unique pole at one other cusp. We can calculate $C\_0$ using the formulas above: If $\zeta\_{2N}=e^{2\pi i \frac{1}{2N}}$, then the constant is given by $\left(\frac{\zeta\_{2N}-\zeta\_{2N}^{-1}}{\zeta\_{2N}^{a}-\zeta\_{2N}^{-a}}\right)^m$. 


Since $A\_N$ is a hauptmodl, its image under the Fricke involution is a rational function in $A\_N$. Since it now has a unique $0$ at the cusp $0$ (Where $A\_N$ has the constant $C\_0$) and the constant $C\_0$ at the cusp infinity (Where $A\_N$ has a zero), we must have 
$$
A\_N|\_0\begin{pmatrix}0&-1\\N&0\end{pmatrix}=\frac{-A\_N+C\_0}{X\cdot A\_N+1},
$$
for some $X$, determined by the movement of the pole. If the pole does not move, then $X=0$. Otherwise, $-1/X$ is the constant of $A\_N$ at the cusp where $A\_N|\_0\begin{pmatrix}0&-1\\N&0\end{pmatrix}$ has its pole.


All in all, this suggest you might have hope for $N=7,9$ and $10$, which all give genus $0$ groups. Unfortunately The ratios of the type you've described above aren't hauptmoduln. You might be able to find similar results if you allow for larger order products.


