I'm interested in the existence of certain orthogonal transformations over $\mathbb F = \mathrm{GF}(2^t)$: matrices $M: \mathbb F^d \to \mathbb F^d$ for which $M^{\!\!\;\mathsf T\!\!\;} M = I_d$ over $\mathbb F$. Specifically: let $d \geqslant 2$ be even, and $\mathbf v \in \mathbb F^d$ be such that $\mathbf v \cdot \mathbf v = 1$, where $\mathbf a \cdot \mathbf b$ is the dot product. Is there an orthogonal matrix $M: \mathbb F^d \to \mathbb F^d$ whose first column is $\mathbf v$?
If $d > 1$ is odd, we cannot do this for all $\mathbf v$: for instance, if $\mathbf v = [\, 1 \;\; 1 \;\; \cdots \;\; 1\,]^{\mathsf T}$ for $d$ odd, all vectors which are orthogonal to $\mathbf v$ are also orthogonal to themselves.
If $d = 2^m$, we can construct such an $M$ by taking the columns of the matrix to differ from $\mathbf v$ by transpositions of the coefficients: index the coefficients of $\mathbf v$ by boolean strings $x \in \{0,1\}^m$; and define $\mathbf w^{(z)} = \sum_x v_x \!\;\mathbf e_{x \oplus z}$ for each $z \in \{0,1\}^m$. We then have $\mathbf w^{(y)} \cdot \mathbf w^{(z)} = \delta_{y,z}$ in $\mathbb F$.
But for $d$ not a power of $2$ it is not obvious how to construct such an orthogonal matrix. Do there exist such orthogonal matrices over $\mathbb F = \mathrm{GF}(2^t)$, for all $d,t \geqslant 1$ and all such $\mathbf v$, provided $d$ is even?