I think that if $A\vec{x} = \vec{b}$, then $A^{T}\vec{b} = \lambda \vec{x}$. I think this because any matrix $A$ can be decomposed into
$$ A = rotation1 \leftarrow scaling \leftarrow rotation2$$
and $A^{-1}$ I think should be
$$ A^{-1} = rotation2^{-1}\leftarrow scaling^{-1}\leftarrow rotation1^{-1}$$
and $A^{T}$ should also be very similar
$$A^{T} = rotation2^{-1} \leftarrow scaling \leftarrow rotation1^{-1}$$
Now, the transpose and the inverse only differ in scale. So I think that $$ if\quad A^{-1}\vec{b} = \vec{x}, \quad\quad then \quad A^{T}\vec{b} = \lambda\vec{x}$$
Now, I have just started learning about the four fundamental spaces of a matrix, and I am confused about what should be the geometric meaning of $A^{T}$.
My intuition is made up of the following image.
If you could tell me where my intuition is wrong in the above equations and the essence of the transpose of a matrix, then that will be of great help. Thanks...