Bulgaria

Let $n=p^k{q}^l$, where $p<q$ are prime numbers, $k,l\in \mathbb{N}$ and let $u=\phi (\tau (n)),v=\tau (\phi (n))$. We have that $u=\phi ((k+1)(l+1))$ and $v=\tau (p^{k-1}{q}^{l-1}(p-1)(q-1))$. Obviously $u<kl+k+l$ and $$v\ge \tau (p^{k-1}{q}^{l-1}(p-1))+1=kl\tau (p-1)+1.$$ If $p>2$, then $\tau (p-1)\ge 2$ and $v\ge 2kl+1\ge kl+k+l$ ($⟺(k-1)(l-1)\ge 0$), hence $v>u$. Therefore $p=2$ and thus $v=\tau (2^{k-1}{q}^{l-1}(q-1))$. If $m>2$ is a prime and $m|q-1$, then $2m|q-1$. So, $$v\ge \tau (2^k{q}^{l-1}m)=2(k+1)l>2kl+1$$ and again $v>u$. Therefore $q=2^s+1,s\in \mathbb{N}$. It follows now that $v=\tau (2^{k+s-1}{q}^{l-1})=(k+s)l$ and the equality $u=v$ implies $$\phi ((k+1)(l+1))=(k+s)l.$$ It follows from the formula for $\phi$ that if $a>1$ and $b>1$, then $\frac{\phi (ab)}{ab}\le \frac{\phi (b)}{b}$. i.e. $\phi (ab)\le a\phi (b)$ and equality holds only if all prime factors of $a$ divide $b$. When $a=k+1$ and $b=l+1$ we have $$\phi ((k+1)(l+1))\le (k+1)\phi (l+1)\le (k+s)l.$$ Moreover the equality holds if and only if $s=1$, i.e. $q=3$. $\phi (l+1)=l$. Hence $l+1$ is a prime number and all prime factors of $k+1$ divide $l+1=r$, i.e. $k+1=r^t,t\in \mathbb{N}$. Therefore the desired numbers are all integers of the form $n=2^{r^t-1}{3}^{r-1}$, where $r$ is a prime number and $t\in \mathbb{N}$.