Does the "nothing" quantifier commute variable-wise? What about the unique existential quantifier?

Question

Let $N$ denote the "for no thing" quantifier, and let $I$ denote the "there exists a unique thing" quantifier. Suppose we have the formula $Rxy$, where $R$ is a binary relation symbol. Are the formulas $(Nx)(Ny)Rxy$ and $(Ny)(Nx)Rxy$ equivalent? What about the formulas $(Ix)(Iy)Rxy$ and $(Iy)(Ix)Rxy$? I know that both the universal and existential quantifiers commute variable-wise.

Answer 1

If $Nx$ is defined as $\neg \exists x$, then
$(Nx)(Ny)Rxy=\neg(\exists x, \neg(\exists y, Rxy))$
$= \forall x, \neg \neg (\exists y, Rxy)$
$= \forall x, \exists y, Rxy$
So this is different from $(Ny)(Nx)Rxy = \forall y, \exists x, Rxy$

For $(Ix)(Iy)Rxy$, take function $f(x)=x^2$ on $\mathbb{R}$.
$(Iy)(Ix), y=f(x)$ is true: there is only one $y$ for which there is only one $x$ such that $y=f(x)$, that's $x=y=0$.
However $(Ix)(Iy), y=f(x)$ is false: there are plenty of $x$ such that there exists only one $y$ such that $y=f(x)$: this is true for all $x$.

By the way, very nice question. I would have thought that both $N$ and $I$ would commute variable-wise.