I am looking for multiple proofs of that statement: here $\phi(n)$ denotes the Euler’s totient $$\sum_{d|n}{\phi(d)}=n$$

Here’s one:

By unique factorisation theorem: $n=\prod_{k=1}^{m}{p_k^{\alpha_k}}$ and $d=\prod_{k=1}^{m}{p_k^{\beta_k}}$ where $0\leq \beta_k\leq \alpha_k$ so:

\begin{align} \sum_{d|n}{\phi(d)}&=\sum_{0\leq \beta_k\leq \alpha_k}{\phi\left(\prod_{k=1}^{m}{p_k^{\beta_k}}\right)}\\ &= \sum_{0\leq \beta_k\leq \alpha_k}{\prod_{k=1}^{m}\phi({p_k^{\beta_k})}}\\ &=\sum_{0\leq \beta_k\leq \alpha_k}{\prod_{k=1}^{m}{(p_k^{\beta_k}-p_k^{\beta_k-1}})}\\ &=\prod_{k=1}^{m}{\sum_{0\leq \beta_k\leq \alpha_k}{(p_k^{\beta_k}-p_k^{\beta_k-1}}})\\ &= \prod_{k=1}^{m}{p_k^{\alpha_k}}\\ &=n. \end{align}