Note that $\varphi(1)=\varphi(2)$. More generally, $\varphi(2n)=\varphi(n)$ if $n$ is odd.
So the $\varphi$-function is not one to one. It is also not onto. For if $b\gt 1$ is odd, there is no $n$ such that $\varphi(n)=b$.
**Remark:** A fairly recent result of Ford shows that for any integer $k\ge 2$, there is a number $b_k$ such that $\varphi(x)=b_k$ has exactly $k$ solutions.
It is not known whether there is any $b$ such that $\varphi(n)=b$ has exactly $1$ solution. Please see Carmichael's Conjecture for details.
So the $\varphi$-function is quite spectacularly non-invertible.