Yes except eigenvalues need not be 1. e^(kx) is an eigenvector of d/dx for all k but is only a fixed point when k=1

Right, in topology a “point” can represent any mathematical object - think “bound variable”.

> So… fixed points?

Fixed directions. The operator operating on an eigenvector doesn't change the vector's direction, but can change its magnitude (i.e. length). (Except, if you change the length from positive to negative, then you kind of change, or flip, the direction to 180 degrees opposite.)