Let $\Bbb{N}^\Bbb{N}$ be the set of all functions $(x_n\mid n\in\Bbb{N})$ from $\Bbb{N}$ into itself (I identify sequences with their images, as usual). I know this is a metrizable space with compatible metric defined as follows: $d(x,y)=2^{-n-1}$ where $n=\mathrm{inf}\{m\in\Bbb{N}\mid x_m\ne y_m\}$ (clearly $d(x,x)=0$ for every $x\in\Bbb{N}^\Bbb{N}$). Denote by $S(\Bbb{N})$ the group of permutations on $\Bbb{N}$.
In "Classical Descriptive Set Theory" by Kechris, the author states (Birkhoff-Kakutani Theorem (9.1), pp. $58$) that
If $G$ is metrazable, $G$ admits a compatible metric $d$ which is left-invariant: $d(zx,zy)=d(x,y)$.
Here is my question: is the metric $d(x,y)=2^{-n-1}$ defined above left-invariant on $S(\Bbb{N})$?
My attempt: the case $x=y$ is trivial; if $x\ne y$, then I want to prove that $$x_m\ne y_m \iff (zx)_m\ne (zy)_m.$$ This is equivalent to saying that $x_m=y_m \iff (zx)_m=(zy)_m$.
As $(zx)_m=z_{x_m}$, it sufficies to show that $$x_m=y_m \iff z_{x_m}=z_{y_m}.$$ Then $(\implies)$ is clear and the converse follows by the fact that $x,y,z$ are bijections. Am I right?
Thank you to @user647486 for helpful comments.