Yes, this would correspond to a matrix with only one eigenvector. Example: $f(x)=Ax$ with $A=\begin{bmatrix} 2 & 1\\\ 0 & 2 \end{bmatrix}$ or $A=\begin{bmatrix} 2 & 1 & 0 \\\ 0 & 2 & 1 \\\ 0 & 0 & 2 \end{bmatrix}$. The direction of the invariant line passing through the origin is $v=(1,0)^T$ and $v=(1,0,0)^T$, resp (that's the eigenvectors).