Uniqueness of Yang-Mills theory Question:

Is there any sense of uniqueness in Yang-Mills gauge field theories?

Details:
Let's say we are after the most general Lagrangian Quantum Field Theory of (possibly self-interacting) $N$ spin $j=1$ particles (and matter). Yang-Mills' construction is based on the following:


*

*Pick a compact semi-simple Lie Group $G$ with $\dim G=N$, and introduce $N$ vector fields $A_\mu^a$, $a=1,\dots,N$. Then
$$
F^a_{\mu\nu}\equiv 2\partial_{[\mu}A_{\nu]}^a+gf^{abc}A_\mu^b A_\nu^c
$$

*The Lagrangian is given by
$$
\mathcal L=-\frac12\text{tr}(F^2)+\mathcal L_\mathrm{matter}(\psi,\nabla\psi)+\text{gauge-fixing}
$$
where $\nabla\psi\equiv\partial\psi-ig T^a A^a$.
My question is about how unique this procedure is. For example, some questions that come to mind:


*

*Is $-\frac12\text{tr}(F^2)$ the most general Lagrangian $\mathcal L=\mathcal L(A^a_\mu)$ that leads to a consistent theory? or can we add new self interactions, and new free terms, without spoiling unitarity, covariance, or renormalisability?

*Is minimal coupling $\partial\to \nabla$ the most general introduction of interactions with the matter fields? or can we add non-minimal interactions without spoiling unitarity, covariance, or renormalisability?
In short: does Yang-Mills' construction lead to the most general Lagrangian that can accommodate the interactions of these spin $j=1$ particles consistently? This construction has many different ingredients, some of which can be motivated through geometric considerations, but I've never seen any claim about uniqueness.
 A: If you don't impose power-counting renormalizability, there are a host of other possibilities, since higher order derivatives or higher order interactions can be introduced. For example, terms $(Tr(F^2)^m)^n$ and are gauge invariant but for $m>1$ or $n>1$ not renormalizable. 
If you impose power-counting renormalizability, uniqueness is fairly straightforward up to trivial field transformations. To see this one first looks at monomials - products of fields and their derivatives. By renormalizability, the total degree is not allowed to be bigger than 4. Each partial derivative $d_j=\partial_j$ counts as degree 1, each Bose fields $A_j$ as degree 1, and each Fermion field $\psi_j$ as degree 3/2. Moreover, Fermions must appear an even number of times to yield a scalar Lagrangian. This leads to a quite short list of possibilities: Up to 4 $A$s and $d$s, or $\psi\psi, d\psi\psi, A\psi\psi$, all with all possible indices. The general renormalizable local Lagrangian density is a linear combination of these, at fixed $x$. Now impose Poincare invariance and gauge invariance, and the only linear combinations left are the ones that one sees everywhere. For a single Yang-Mills field and nothing else (i.e., your question in the narrow sense) 
the only freedom left is rescaling the fields, which eliminates an arbitrary factor in front of the trace. In the presence of fermionic fields there is the additional freedom to take linear combinations of Fermionic fields as new fields, which may be used to reduce the associated bilinear forms to weighted sums of squares. 
If one drops gauge invariance, there are lots of other possible Langangian densities, for example a mass term, products of it with the terms described, and even more. 
Note that proving the renormalizability of nonabelian gauge theories with broken symmetries was a highly nontrivial achievement (around hundred pages of published argument) worthy of a nobel prize for Veltman nd 't Hooft. Thus it is unreasonable to explain in an answer the reasons for where precisely the borderline is bewteen renormalizable and nonrenormalizable.
The answer to your question, ''Maybe I can put my question in simpler terms: is there any room for modifications in the Standard Model without introducing new fields? can we add new interactions among the gauge bosons (W,Z,…) and/or the matter fields without spoiling unitarity, covariance, or renormalisability? (at the perturbative level at least; here I dont care about θ terms, etc.)'' related to the bounty (which will disappear in a few hours) 
is no, essentially by an extension of the above reasoning (including the 100 pages of proof of renormalizability).
