I'm looking for an IFS acting on $\mathbb{R}^2$ whose components are affine transformations, e.g.
$\vec{f}(\vec{x})=A\vec{x}+\vec{b}$
where $A\in GL_2(\mathbb{R})$ and $\vec{x},\vec{b}\in\mathbb{R}^2$. This IFS must have as a fixed set some finite 1-D segment of $\mathbb{R}^2$. It is clear that degenerate IFSs exist that do this. For example, the two-component IFS
$f_1\left(\begin{bmatrix} x\\ y\end{bmatrix}\right) = \begin{bmatrix}1/2& 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}+\begin{bmatrix}0 \\ 1\end{bmatrix}$
$f_2\left(\begin{bmatrix} x\\ y\end{bmatrix}\right) = \begin{bmatrix}-1/2& 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}+\begin{bmatrix}1/2 \\ 1\end{bmatrix}$
has the segment from $(0,1)$ to $(1/2,1)$ as a fixed point. Is it possible to get the same fixed point with a nondegenerate/invertible transformation?
Let $G = \left\{g_1, g_2\right\}$ where each $g_k$ is invertible:
$$ \begin{aligned} g_1(v) &= \frac{1}{2} v \\ g_2(v) &= \frac{1}{2} v + \left[\frac{1}{2} \quad 0\right]^T \end{aligned} $$
Then $G$ has fixed point a segment $S = \left[\left[0 \quad 0\right]^T, \left[1\quad0\right]^T\right]$
Now let $h$ be an invertible affine transformation mapping $S$ to the segment in the question. Finally conjugate each $g_k$ by $h$ to give the desired functions: $$ f_1 = h \circ g_1 \circ h^{-1} \\ f_2 = h \circ g_2 \circ h^{-1} \\ $$