Integers whose totient divides them ($\phi (n)\mid n⟺n=1\text{ or }n=2^c{3}^d,c\ge 1,d\ge 0$)

If $n>1$ is odd, its Euler totient function is even and cannot divide $n$. Otherwise write $n=2^c{\prod }_{j=1}^r{p}_j^{a_j}$ with odd primes $p_j$ and $c\ge 1$. Comparing powers of two in $\phi (n)\mid n$ gives ${\sum }_j{v}_2(p_j-1)\le 1$. Thus there is at most one odd prime $p$, with $v_2(p-1)=1$. Since $p-1$ divides $\phi (n)$, it divides $n$, and is coprime to $p$; with only the prime factors $2,p$ available it must be a power of two. Hence $p-1=2$ and $p=3$. Conversely, for $c\ge 1$ the prime-power totient formula verifies divisibility for $2^c{3}^d$, and $n=1$ is immediate.