To get the generating function of $F_n$, I use the basic recurrence $F_n = F_{n-1} + F_{n-2}$.
I think the same approach will work for $F_{kn}$. For example,
$$F_{2n} = 3F_{2(n-1)} - F_{2(n-2)}$$
$$F_{3n} = 4F_{3(n-1)} + F_{3(n-2)}$$
from A001519 and A014445. I conjecture for any $k$: $$F_{kn} = aF_{k(n-1)} + bF_{k(n-2)}$$
How can I find $a$ and $b$? My ideas were to try Binet's Formula (though that requires dealing with irrational numbers) and the matrix form
$$A^n = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n = \begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix}$$ so that $A^{kn} = aA^{k(n-1)} + bA^{k(n-2)} $.
Since $A$ is invertible this is equivalent to $A^{2k} = aA^k + bI$. Maybe this can be solved from here?
Binet's formula states that $F_n=\alpha\tau^n+\beta\tau'^n$ where $\tau=\frac12(1+\sqrt5)$ and $\tau'=\frac12(-1+\sqrt5)$. Here it's immaterial what $\alpha $ and $\beta $ are.
Therefore $$F_{kn}=\alpha(\tau^k)^n+\beta(\tau'^k)^n.$$ The recurrence for $F_{kn}$ has characteristic equation $$(X-\tau^k)(X-\tau'^k)=X^2-(\tau^k+\tau'^k)X+\tau^k\tau'^k.$$ But $\tau^k\tau'^k=(-1)^k$ and $\tau^k+\tau'^k=L_k$, the $k$-th Lucas number. ($L_0=2$, $L_1=1$, $L_n=L_{n-1}+L_{n-2}$).
Therefore $$F_{kn}=L_kF_{k(n-1)}-(-1)^kF_{k(n-2)}.$$