In this post, I had an idea that
$$\varphi(n) = \sum_{d \mid n} d \cdot \mu(\frac{n}{d})$$
has this property, where $\mu$ is the Mobius function.
Let $\star$ be the Dirichlet convolution of functions $f,g: \mathbb{N} \to \mathbb{C}$ defined by $$(f \star g)(n) = \sum_{ab = n} f(a)g(b) = \sum_{d \mid n} f(d)g(\frac{n}{d})$$
If $\text{Id}$ is the identity function (i.e., $\text{Id}(n) = n$ for all $n \in \mathbb{N}$), $1$ is the "$1$" function (i.e., $1(n) = 1$ for all $n \in \mathbb{N}$), and $\varphi$ to be the Euler's totient function, is the Euler's totient function the unique function that satisfy $\text{Id} = 1 \star \varphi$?
Indeed! I made a proof.
Proof:
Notice, first, that for any two functions $f,g: \mathbb{N} \to \mathbb{C}$
$$ f = 1 \star g \iff g = f \star \mu$$
and $$\varepsilon = 1 \star \mu$$
where $\varepsilon(n) = \lfloor \frac{1}{n} \rfloor$ for all $n \in \mathbb{N}$, and $\mu$ is the Mobius function.
Denote $D := \{f: \mathbb{N} \to \mathbb{C} \, \mid \, f(1) \ne 0\}$. Then $(D, \star)$ is an abelian group. The identity in $D$ is $\varepsilon$. Also, each functions $f$ have Dirichlet inverses iff $f(1) \ne 0$. That is, $(\forall f \in D)(\exists! g \in D)$ such that $$f \star g = \varepsilon \iff f(1) \ne 0$$.
Notice, $\varphi \in D$. So, if $\text{Id} = 1 \star f = 1 \star \varphi$, then,
$$\begin{equation*} \begin{split} \text{Id} = 1 \star f = 1 \star \varphi &&\iff &\mu \star 1 \star f = \mu \star 1 \star \varphi\\ &&\iff &(\mu \star 1) \star f = (\mu \star 1) \star \varphi\\ &&\iff &\varepsilon \star f = \varepsilon \star \varphi\\ &&\iff &f = \varphi &\square \end{split} \end{equation*}$$
Question
Is this proof correct and valid?
Edit
I searched this in this link.