What is the difference between
a) $\mathbf{f} \rightarrow \mathbf{f}(\mathbf{R}^T\cdot\mathbf{x})=\mathbf{f}(\mathbf{x}^\prime)$, (Wikipedia, the article about rotational invariance)
and
b) $\mathbf{f} \rightarrow \mathbf{R}\cdot \mathbf{f}(\mathbf{R}^T\cdot\mathbf{x})=\mathbf{f}^\prime(\mathbf{x}^\prime)$, (also from Wikipedia, cannot find the reference right now, it was a certain article explaining rotational invariance)
Both cases are supposed to show rotationally invariant functions. What is the difference between the 2 cases?
Attempt at explanation:
a) I understand this to mean: Given is a vector-function $\mathbf{f}=(f_1,f_2,...,f_n)$, where $f_i = f_i(\mathbf{x})=f_i(x_1,x_2,...,x_n)$. $\mathbf{R}$ is the rotation matrix; when we multiply $\mathbf{R}^T$ by the vector $\mathbf{x}$, we get the rotated vector $\mathbf{x}^\prime$, that is, $\mathbf{R}^T \cdot\mathbf{x} = \mathbf{x}^\prime$. Then the function $\mathbf{f}$ has the same form in both $\mathbf{x}$-coordinates (before rotation) and $\mathbf{x}^\prime$-coordinates (after rotation). Example: $f_i=F_i(x_1^2+x_2^2+...+x_n^2)$ and $f_i=F_i(x_1^{\prime 2}+x_2^{\prime 2}+...+x_n^{\prime 2})$, where $F_i$ are arbitrary functions. We call $\mathbf{f}$ a rotationally-invariant function, because when coordinates are rotated the function does not change only becomes expressed in the new rotated coordinates.
b) This is harder to understand. Again, the argument position-vector $\mathbf{x}$ is rotated; but the function is also rotated (backwards?). Somehow, the function doesn't change?
Could someone help explain this (possibly with examples)?