Hopefully easy Lang-Steinberg computation: for Weyl elements 
How do you write an element of the Weyl group as $g^{-1} F(g)$?

For instance, let $G = \langle x_1(t), x_2(t), x_{-1}(t), x_{-2}(t) : t \in K \rangle$ where $K$ is an algebraically closed field of characteristic $p$, where $F:G\to G$ is the group homomorphism sending $x_i(t)$ to $x_i(t^q)$ for some power $q$ of $p$, and where $x_i(t)$ are given as follows:
$$
x_1(t) = \begin{bmatrix}1 & t & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -t \\ 0 & 0 & 0 & 1 \end{bmatrix}, \quad
x_2(t) = \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & t & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}
$$
with $x_{-i}(t)$ the transpose of $x_i(t)$. $G$ is (affine, simple, connected) linear algebraic group $\operatorname{Sp}_4(K) = B_2(K)$, and $G^F$ is the finite group of Lie type $\operatorname{Sp}_4(q)$, both preserving the alternating form $J=\left[\begin{smallmatrix}0&0&0&1\\0&0&1&0\\0&-1&0&0\\-1&0&0&0\end{smallmatrix}\right]$.
Let $n_i(t) = x_i(t) \cdot x_{-i}(-1/t) \cdot x_i(t)$ and $w_i = n_i(1)$. Explicitly, $$
w_1 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{bmatrix}, \quad
w_2 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}
$$

How do you write $w_i$ as $g^{-1} F(g)$ for $g \in G$?

I would like an answer that works independently of $q$, though I am fine with restricting to $q$ odd.
I know such a $g$ exists by Lang (or Lang-Steinberg), but I'm not sure how to find it. I've seen papers that give Las Vegas polynomial algorithms to do this, but I suspect for such “nice” elements, the output can be chosen to be nice, rather than random.
 A: I suspect that a solution in $SL_2$ could be made to work in any "single positive root" version, i.e. in a subgroup like $SL_{2,\alpha}=\langle x_\alpha(t),x_{-\alpha}(t)\mid t\in k\rangle$. So let's try $SL_2(k)$. The equation $g^{-1}F(g)=w$ is equivalent to $F(g)=gw$.
In the case of $SL_2$ we have
$$
w=\left(\begin{array}{rr}0&1\\-1&0\end{array}\right).
$$
There we want to find 
$$
g=\left(\begin{array}{rr}a&b\\c&d\end{array}\right)
$$
such that
$$
\left(\begin{array}{rr}a^q&b^q\\c^q&d^q\end{array}\right)
=F(g)=gw=\left(\begin{array}{rr}-b&a\\-d&c\end{array}\right).
$$
This implies the equations $a=b^q=(-a^q)^q=-a^{q^2}$, and similarly $c=d^q=-c^{q^2}$.
Therefore $a,b,c,d$ are all negated by the second power of Frobenius (assuming $q$ odd here). But repeating the dose shows that they are fixed points of $F^4$. This implies that $a,b,c,d\in \Bbb{F}_{q^4}\setminus \Bbb{F}_{q^2}$. Therefore a solution that does not depend on $q$ cannot exist.
Here we need the determinant constraint
$$
1=\det g=ad-bc=ad+a^qd^q
$$
to hold.
Let's write $u=ad$. The constraint becomes $1=u+u^q$. This holds for example when $u=1/2$ for as an element of the prime field $1/2$ is a fixed point of $F$. This suggests the following recipe for $g$. Let $\varepsilon\in\Bbb{F}_{q^4}$ be an element negated under $F^2$. For example we can let $\varepsilon$ be a square root of a non-square in $\Bbb{F}_{q^2}$. Then we can choose
$$
\left\{
\begin{array}{ccr}
a&=&\varepsilon,\\
b&=&-\varepsilon^q,\\
d&=&1/2\varepsilon,\\
c&=&1/2\varepsilon^q.
\end{array}
\right.
$$
