Just a comment: these objects are called arcs and have been well studied in projective geometry. See en.wikipedia.org/wiki/Arc_(projective_geometry).

Maybe it is worth mentioning here that this is only a special instance of Wedderburn's Theorem stating that any central simple algebra is isomorphic to a matrix algebra over a division ring. In that sense, it is perhaps not the kind of example the OP is looking for.

@Echo Sure, now that you have rephrased it as "epimorphism in the category of rings" and added the example, I of course agree. (From en.wikipedia.org/wiki/Epimorphism#Terminology: "Saunders Mac Lane attempted to create a distinction between epimorphisms, which were maps in a concrete category whose underlying set maps were surjective, and epic morphisms, which are epimorphisms in the modern sense. However, this distinction never caught on.")

Wait, didn't you write "ring epimorphism"? (Perhaps you meant "epic morphism" then, which isn't the same!)

It's called "quotients of $\mathbb{Z}[X_1,\dots,X_n]$"...

@JacksonWalters Of course, $\langle T_r, T_t \rangle$ is still a group, because it's a subgroup of $\operatorname{Sym}(\mathbb{F}_{2^n})$. To answer the question about "how non-linear" the map $T$ is, you could try intersecting the image of $T$ with $\mathrm{GL}_n(2)$ and computing its index in $\operatorname{im}(T)$, but I'm not sure what sort of information you would hope to get out of this.

@JacksonWalters Sorry, but I still disagree. Any map from a vector space to another vector space maps basis elements to a linear combination of basis elements, even if it is non-linear. In fact, looking at the reference by Brown and Loehr you provide, they do quite a bit of work to show linearity, so you can't expect a one-line proof. Even more: I checked this for $p=31$, with $g(x) = x^5 + x^2 + 1$, and it turns out to be false: $T_r$ is not linear. (If you want, I can write down more details in an answer.)

@JacksonWalters I don't see why $T_f$ is linear (and I certainly don't understand your short argument). For instance, why is $T_r$ linear? Could you elaborate?

Do you mean that you consider $S_n$ as a Coxeter group with generating set $s_1 = (12)$, $s_2 = (23)$, $s_3 = (34)$, etc.? What do you mean by a "single commutation"? Do you mean that you replace a single instance of $s_i s_j$ by $s_j s_i$ where $s_i$ and $s_j$ are commuting generators?

Here is your tool: Two diagonal matrices are conjugate to each other if and only if the elements on the diagonal are permutations of each other (because these are the eigenvalues of these matrices, hence invariant under conjugation).

What is your definition of $\mathbf{PSL}(V)$ when $V$ is infinite-dimensional?

Just a comment that might be helpful: since the matrices are $2 \times 2$, we have the equality $\det(A+B)=\det A+\det B+\det A\,\cdot \mathrm{Tr}(A^{-1}B)$ when $A$ is invertible. In particular, if you want $\det(A+B) = \det(A) + \det(B)$, then you need $\mathrm{Tr}(A^{-1}B) = 0$.