Taking cyclic subgroups seems promising for obtaining identities similar to the ones of Nijenhuis, and so I have looked at products of ${\Gamma\left(\dfrac{a^k }{N}\right)}/{\sqrt{2\pi}}$ with $k$ running over a certain range. The factor $\sqrt{2\pi}$ is motivated by the fact that

- in Nijenhuis' formula as well as in the multiplication formula and even in the reflection formula, a factor $\sqrt{\pi}=\Gamma(\frac12)$ occurs with the same multiplicity as the gamma factors (so my products can be considered in fact as "well-poised" gamma quotients),
- without including $\sqrt{2}$, those products, whenever they are integers, have a relatively big 2-valuation (see the factor $2^{b(A)}$ in Nijenhuis' formula $$\prod_{x\in A}{\Gamma\left(\frac{x }{2n}\right)}=2^{b(A)} \sqrt{\pi}^{\,\nu(n)},$$ where $A$ is the subgroup of the multiplicative group $\mathbb Z_{2n}^*$ generated by $n + 2$ or any one of its cosets, $\nu(n)$ its order and $b(A)$ the number of elements in $A$ that are larger than $n$).

For brevity, define for coprime integers $a, N$ $$g(a,N):=\prod_{k=1}^{ind_a(N)} \frac{\Gamma\left(\dfrac{a^k \pmod N}{N}\right)}{\sqrt{2\pi}}.$$ Thus the product is taken over the subgroup $\langle a\rangle$ of $\mathbb Z_N^*$ generated by $a$, with all the arguments of the gamma factors between $0$ and $1$. For an integer $b$ coprime to $N$, define similarly the product over the corresponding coset $b\langle a\rangle=\langle a\rangle b$ as $$g_b(a,N):=\prod_{k=1}^{ind_a(N)} \frac{\Gamma\left(\dfrac{a^kb \pmod N}{N}\right)} {\sqrt{2\pi}}.$$

With this notation, the initial identity reads $g(9,14)=\sqrt{2}$, and we further have the somewhat complementary one $g_3(9,14)=1/\sqrt{2}$.

Once you know where to (ask a computer to) search, it is easy to find, in a few minutes, dozens of those products that appear numerically to be algebraic.

I have excluded the self-complementary groups, i.e. the groups for which $\langle a\rangle=\langle N-a\rangle $, because for those, we already know that $g(a,N)$ is algebraic by the reflection formula. Likewise I have excluded subgroups whose members form essentially an arithmetic sequence (more precisely, a set $\{\lambda s+t\}\cap \mathbb Z_N^*$ for given $s,t$), as those can be handled by the multiplication formula and yield again algebraic products. Call the remaining subgroups and the associated gamma products non-trivial.

A systematic search (with reasonably high numerical precision) shows that of the non-trivial algebraic gamma products, most are of the form $q^{u/v}$ with integers $u,v$. Generally, $u/v$ is positive and $q$ a prime dividing $N$. But there are exceptions like $g(24,203)=7^{-1/2}$ or $g(103,420)=(3^3\cdot5^3\cdot7)^{1/6}$.

For $n\le300$, there are $106$ non-trivial algebraic gamma products, $88$ of which can be written in form $q^{u/v}$. (With $v=1$ for 36 of them.) The $18$ remaining ones occur for $N= 60,105,120,140,156,180,220,231,255,285,300$ (note that all these $N$'s have at least three prime divisors) and have minimal polynomials of degrees $2,4,6$ or $8$. The latter holds for all $N\le 1000$.

Moreover so far all of them can be written with radicals, e.g. $g(103,105)=\sqrt[3]{3(1701+166\sqrt{105})}$ or $g(41,156)=2 \sqrt[4]{13}(2\sqrt{3}+\sqrt{13})$ or $g(83,120)=\sqrt{3(\sqrt{3}-1)(\sqrt{30}-5)}$.

A few of these identities, e.g. the one for $g(83,120)$, can be derived by using the standard formulas (thus in the way the OP asks next to the end). But I doubt this is possible where e.g. $\sqrt{13}$ occurs.

You can get more similar identities by taking the cosets of the same groups. I haven't looked systematically at them, but conjecturally, if $g(a,N)$ is algebraic, so is $g_b(a,N)$. Note that generally, $g_b(a,N)/g(a,N)$ does not seem to be algebraic.

The distribution of the non trivial subgroups seems to be at least as irregular as the distribution of the primes and not even very much correlated with the structure of $\mathbb Z_N^*$.