Let $F_n^{(k)}(x)= \sum_j {\binom{n+(k-1)j}{kj} x^j}$ and $G_n^{(k)}(x)= \sum_j {\binom{n+j}{kj} x^j}.$

I am interested in the coefficients ${a_{n,k,j}}$ such that $$G_n^{(k)}(x)=\sum_{j\geq0 }{a_{n,k,j}} F_j^{(k)}((-1)^ { k}x).$$ Computations suggest the following:

Let $z\sum_{j\geq 0}C_{1,j}^{(k)}z^j$ be the inverse series of $\sum_{j=1}^k (-1)^{j-1}z^j$ and let $(\sum_{j\geq 0}C_{1,j}^{(k)}z^j)^m=\sum_{j\geq 0}C_{m,j}^{(k)}z^j.$ Then $$ a_{n,k,j}=(-1)^{k j}C_{kj+1,n-(k-1)j}^{(k-1)}.$$ Any idea how to prove this?

**Remark:** $C_{1,n}^{(2)}=C_n$ are the Catalan numbers and

$( C_{1,n}^{(3)})_{n\geq 0}=(1,1,1,0,-4,-14,-30,-33,\dots).$ (cf. OEIS, A103779).

**Edit:** Perhaps the following observation may be useful.
A matrix inversion theorem of Gould and Hsu implies a similar result:
The coefficients $c_{n,k,j}$ which give $$\sum_j {c_{n,k,j}} F_j^{(k)}(x)=x^n$$ are $(-1)^{n-j} A_{n-j,k,kj+1}$ where $A_{n,k,r}=\frac{r}{kn+r}\binom{kn+r}{n}$ are Fuss-Catalan numbers.
Here we have $\sum_{n}A_{n,k,r}x^n=(\sum_{n}A_{n,k,1}x^n)^r $ and $ x\sum_{n}A_{n,k,1}x^{(k-1)n}$ is the inverse series of $y-y^k.$
Is there a combinatorial or other reason for the appearance of these special inverse series?