The golden ratio $\phi$ for $_2F_1\big(\frac16,\frac16,\frac23,-2^7\phi^9\big)$ and $_2F_1\big(\frac16,\frac56,1,\frac{\phi^{-5}}{5\sqrt5}\big)$?

Question

I. Context

Given the golden ratio $\phi$, then we have the nice closed-forms,

\begin{align} _2F_1\left(\frac16,\frac16,\frac23,-2^7\phi^9\right) &= \frac{3}{5^{5/6}}\phi^{-1}\\[6pt] _2F_1\left(\frac16,\frac56,1,\,\frac{\phi^{-5}}{5\sqrt5}\right)\, &= \left(\frac{3\sqrt3}{8}\right)^{1/3}\left(\frac{5\sqrt5}{4}\right)^{1/6}\,\frac{\Gamma\big(\tfrac13\big)^3}{2\pi^2} \end{align}

The fact the $_2F_1\big(a,b;c;z_k\big)$ have simple relations between their $(a,b,c)$, the parameter $z_k$ involves the same constant (the golden ratio $\phi$), and they have closed-forms is suggestive they have a connection.

In fact, given the j-function $j(\tau)$, the $z_k$ were derived similarly.

$$\frac{j(\tau_1)}{1728} = 4z_1(1-z_1)\tag1$$

$$\frac{j(\tau_2)}{1728} = \frac1{4z_2(1-z_2)}\tag2$$

The parameter $z_1$ was discussed in an old MSE post, while $z_2$ (the case $s=\frac16$) was moved from a new MO post as it was getting too long.

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II. Question

Solving the quadratic for $z_2$ in $(2)$,

$$\delta =z_2= \frac12\left(1\color{red}-\sqrt{1-\frac{1728}{j(\tau)}}\right)$$

Is it true that $j(\tau)$ is a $24$th power,

$$j(\tau) \overset{\color{red}?}= \left(\frac{\sqrt{_2F_1\big(\tfrac16,\tfrac56,1,\delta\big)}}{\eta(\tau)}\right)^{24}$$

for $\tau = \sqrt{-d}\,$ or $\tau = \frac{1+\sqrt{-d}}2\,$ for integer $d>0$ and $d>3$, respectively?

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III. Data

If true, the formula gives closed-forms for infinitely many hypergeometrics $_2F_1$ using the gamma function,

1. Let $\tau = \sqrt{-3}$, so $j(\tau)=2\times30^3$. We have the one from the title, $$_2F_1\left(\frac16,\frac56,1,\,\frac{\phi^{-5}}{5\sqrt5}\right) = \left(\frac{3\sqrt3}{8}\right)^{1/3}\left(\frac{5\sqrt5}{4}\right)^{1/6}\,\frac{\Gamma\big(\tfrac13\big)^3}{2\pi^2}$$
2. Let $\tau = \sqrt{-7}$, so $j(\tau)=255^3$. Then, $$_2F_1\left(\frac16,\frac56,1,\tfrac12\Big(1-\sqrt{1-\tfrac{1728}{255^3}}\Big)\right) = \left(\frac{255}7\right)^{1/4}\,\frac{\Gamma\big(\frac17\big) \Gamma\big(\frac27\big) \Gamma\big(\frac47\big)}{2\times(2\pi)^2\;}$$
3. Let $\tau = \frac{1+\sqrt{-7}}2$, so $j(\tau)=-15^3$. Then, $$_2F_1\left(\frac16,\frac56,1,\tfrac12\Big(1-\sqrt{1-\tfrac{1728}{-15^3}}\Big)\right) = \left(\frac{15}7\right)^{1/4}\,\frac{\Gamma\big(\frac17\big) \Gamma\big(\frac27\big) \Gamma\big(\frac47\big)}{(2\pi)^2}$$
4. Let $\tau = \frac{1+\sqrt{-11}}2$, so $j(\tau)=-32^3$. Then,

$$_2F_1\left(\frac16,\frac56,1,\tfrac12\Big(1-\sqrt{1-\tfrac{1728}{-32^3}}\Big)\right) = \left(\frac{32}{11}\right)^{1/4}\,\frac{\Gamma\big(\frac1{11}\big) \Gamma\big(\frac3{11}\big) \Gamma\big(\frac4{11}\big) \Gamma\big(\frac5{11}\big) \Gamma\big(\frac9{11}\big)}{(2\pi)^3}$$

and so on for other $\tau$.