What about rotation by 180 degrees? In Sheldon Axler's Linear Algebra on page 76 he writes:
The rotation of a nonzero vector in $R^2$ is obviously never equals a scalar multiple of itself. 
Why is rotation by 180 degrees not the same as multiplication by $-1$? 
 A: The context of that sentence is:

For a more complicated example, consider the operator $T\in\mathcal L(\mathbf F^2)$ defined by
  $$T(w,x)=(-x,w).$$
  If $\mathbf F=\mathbf R$, then this operator has a nice geometric interpretation: $T$ is just a counterclockwise rotation by $90^\circ$ about the origin in $\mathbf R^2$. An operator has an eigenvalue if and only if there exists a nonzero vector in its domain that gets sent by the operator to a scalar multiple of itself. The rotation of a nonzero vector in $\mathbf R^2$ obviously never equals a scalar multiple of itself. Conclusion: if $\mathbf F=\mathbf R$, the operator $T$ defined by 5.4 has no eigenvalues. However, if $\mathbf F=\mathbf C$, the story changes. To find the eigenvalues of $T$, [...]

I think it's reasonable to conclude that the sentence in question is referring specifically to $T$ by "the rotation",  not a general rotation by any angle.
That said, you're right! Rotation by 180 degrees in the plane is the same as multiplication by $-1$. For that matter, rotation by 360 degrees is the same as multiplication by $1$.
