By request, here's the integer linear programming formulation I used. Let $C$ be the set of $36$ cells that are not blocked by rooks. Let $t$ be the target number of steps, and let $S=\{0,\dots,t\}$ be the set of steps. Let $D=\{(d_i,d_j)\in \mathbb{Z}^2:|d_i d_j| = 2\}$ be the set of eight directions. Let binary decision variable $x_{i,j,s}$ indicate whether cell $(i,j)$ is occupied by a knight at step $s$. Let binary decision variable $y_{d_i,d_j,s}$ indicate whether direction $(d_i,d_j)$ is chosen at step $s$. The problem is to minimize $\sum_{(i,j)\in C} x_{i,j,t}$ subject to
\begin{align}
\sum_{(d_i,d_j) \in D} y_{d_i,d_j,s} &= 1 &&\text{for $s \in S \setminus \{0\}$} \tag1\\
x_{i,j,0} &= 1 && \text{for $(i,j) \in C$} \tag2\\
\sum_{\substack{(d_i,d_j) \in D:\\ (i+d_i,j+d_j) \notin C}} y_{d_i,d_j,s} +
x_{i,j,s-1} - 1 &\le x_{i,j,s} && \text{for $(i,j) \in C, s \in S \setminus \{0\}$} \tag3\\
y_{d_i,d_j,s} + x_{i-d_i,j-d_j,s-1} - 1 &\le x_{i,j,s} && \text{for $(i,j) \in C, s \in S \setminus \{0\}, (d_i,d_j) \in D: (i-d_i,j-d_j) \in C$} \tag4\\
\end{align}
Constraint $(1)$ selects exactly one direction at each step.
Constraint $(2)$ fixes the initial configuration of knights at step $0$.
Constraint $(3)$ enforces the rule that a knight that cannot move in the chosen direction stays in place. That is, $$(y_{d_i,d_j,s} \land x_{i,j,s-1}) \implies x_{i,j,s}.$$
Constraint $(4)$ enforces the rule that a knight that can move in the chosen direction does so. That is,
$$(y_{d_i,d_j,s} \land x_{i-d_i,j-d_j,s-1}) \implies x_{i,j,s}.$$
I solved the problem for each $t$ from $1$ to $16$ to obtain the optimal values
$$26, 19, 14, 11, 9, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 1.$$
So $t=16$ is the minimum number of steps needed to yield one knight.
