Let $q=2^n$ where $n$ is a sufficiently large odd number. Consider the fintie field $\mathbb F_q$ and the symmetric group $\mathrm{Sym}(\mathbb F_q)$ over it.
I use $x^3$ to denote the permutation $x \mapsto x^3$ over $\mathbb F_q$ (It's a permutation because $2 \nmid n$).
My question is: can the set $\left\{x^3,x+b\right\} :=\left\{x^3\right\} \cup \{x+b\}_{b \in \mathbb F_q}$ generating all permutations, i.e. $\mathrm{Sym}(\mathbb F_q)$, by compositions?
By a theorem of Carlitz, $\left\{x^{q-2},ax+b : a \ne 0\right\}$ is a generating set. One can see a discussion here: https://dept.math.lsa.umich.edu/~zieve/papers/carlitz.pdf. The set can be further refined as $\left\{x^{q-2},x+b\right\}$. Proof: set $f(a)=\left(\left(x^{q-2}+a\right)^{q-2}+a^{q-2}\right)^{q-2}+a$. One can check that $f(1)$ is simply $(01)$ and $f(g)$ is cyclic for any $\mathbb F_q^*$ generator $g$.
However, in our case, compositions of $x^3$ can only obtain a subgroup of $\mathbb Z_{q-1}^*$ and might not include $x^{q-2}$ in general.
I also have an independent but related problem: given a constant $t$ and pairs $(x_i,y_i)_{i \in [t]}$ where $x_i \ne x_j, y_i \ne y_j \forall i \ne j$. Consider the permutation polynomial of degree $3^r$ $f_r(x) = (x+b_1)^3 \circ (x+b_2)^3 \circ \cdots \circ (x+b_r)^3$. Then, $\{f(x_i)=y_i\}_{i \in [t]}$ is a system of polynomial equations. Does there exist a constant $r=r(t)$ such that the system has at least one solution over $\mathbb F_q$? I think that this problem is already non-trivial even over $\overline{\mathbb F_q}$.