A tantalizing Gamma quotient to challenge the Rohrlich-Lang Conjecture

A conceptual way to tackle this question is to look at universal distributions on $\mathbf{Q}/\mathbf{Z}$, studied by Kubert and Lang among others. Distributions arise naturally in number theory, see e.g. Lang's article Relations de distributions et exemples classiques.

There is a function $f$ on $\mathbf{Q}/\mathbf{Z}$ with values in an abelian group $A$ which satisfies the distribution relations $\sum_{a \in \mathbf{Z}/N\mathbf{Z}} f(x+a/N)=f(Nx)$ for every $x \in \mathbf{Q}/\mathbf{Z}$ and every $N \geq 1$, and which is universal for this property. The construction of $f$ and $A$ is obvious: take the free abelian group on $\mathbf{Q}/\mathbf{Z}$ and mod out by the distribution relations.

Similarly, you can look at functions on $\frac{1}{N}\mathbf{Z}/\mathbf{Z}$ satisfying the $M$-distribution relations for every $M$ dividing $N$. You get an abelian group $A_N$ and a canonical map $A_N \to A$.

Kubert has shown in the article The universal ordinary distribution that $A$ is a free abelian group, that the map $A_N \to A$ is injective, that $A_N$ is free of rank $\varphi(N)$, and has determined explicit free generators of $A_N$.

The $\Gamma$ function determines a distribution with values in $\mathbf{C}^\times/\overline{\mathbf{Q}}^\times$. This distribution is odd and the Rohrlich-Lang conjecture asserts that it is universal, in other words the induced map $\overline{\Gamma} : A_N^- \to \mathbf{C}^\times/\overline{\mathbf{Q}}^\times$ should be injective, where $A_N^-$ denotes the odd part of $A_N$.

Concretely, you can write your purported relation as a formal $\mathbf{Z}$-linear combination of the $a/N$ with $0 < a <N$ and check whether this divisor belongs to the subspace generated by the $M$-distribution relations for every $M|N$ and by the relations $[x]+[-x]=0$. This is just linear algebra in a $\mathbf{Q}$-vector space of dimension $N$ so should be doable for quite large $N$.

EDIT. In fact, you can simply use a known universal distribution and evaluate it on your divisor. For example, the Bernoulli distribution $$\tilde{B}_1(x) = \sum_{u \in (\mathbf{Z}/N\mathbf{Z})^\times} B_1(ux) [u]$$ taking values in $\mathbf{Q}[(\mathbf{Z}/N\mathbf{Z})^\times]$ is an odd universal distribution. Here $B_1(y) = \{y\}-\frac12$ for every $y \in \mathbf{R}/\mathbf{Z}$ is the Bernoulli polynomial. So you can simply check whether your divisor is in the kernel of $\tilde{B}_1$.

It turns out that triplication is not needed here: the recursion $\Gamma(z+1) = z \Gamma(z)$, the reflection formula $$ \Gamma(z) \Gamma(1-z) = \frac\pi{\sin \pi z}, $$ and the duplication formula $$ \Gamma(z) \Gamma\bigl(z+\frac12\bigr) = 2^{1-2z} \sqrt\pi \Gamma(2z) $$ suffice to reduce the gamma product to a trigonometric identity.

Apply duplication to $z = \{2^n/39\}$ with $0 \leq n < 12$ (where $\{\cdot\}$ is the fractional part) and multiply the resulting $12$ identities, using the recursion when $z>1/2$ to write $\Gamma(z+\frac12)$ and $\Gamma(2z)$ as rational multiples of $\Gamma(z-\frac12)$ and $\Gamma(2z-1)$. Then the factors $\Gamma(\{2^n/39\})$ cancel out. For the remaining factors $\Gamma(\{ \frac{2^n}{39} + \frac12 \})$, apply reflection when $n$ is odd, we get a formula for $A/B$.