Let $n = p^k q^l$, where $p < q$ are prime numbers, $k, l \in \mathbb{N}$ and let $u = \varphi(\tau(n)), v = \tau(\varphi(n))$. We have that $u = \varphi((k+1)(l+1))$ and $v = \tau(p^{k-1} q^{l-1} (p-1)(q-1))$. Obviously $u < kl + k + l$ and
$$v \geq \tau(p^{k-1} q^{l-1} (p-1)) + 1 = kl \tau(p-1) + 1.$$
If $p > 2$, then $\tau(p-1) \geq 2$ and $v \geq 2kl + 1 \geq kl + k + l$ ($\iff (k-1)(l-1) \geq 0$), 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,
$$v \geq \tau(2^k q^{l-1} m) = 2(k+1)l > 2kl + 1$$
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
$$\varphi((k+1)(l+1)) = (k+s)l.$$
It follows from the formula for $\varphi$ that if $a > 1$ and $b > 1$, then $\frac{\varphi(ab)}{ab} \le \frac{\varphi(b)}{b}$.
i.e. $\varphi(ab) \le a \varphi(b)$ and equality holds only if all prime factors of $a$ divide $b$. When $a = k + 1$ and $b = l + 1$ we have
$$\varphi((k+1)(l+1)) \le (k+1) \varphi(l+1) \le (k+s)l.$$
Moreover the equality holds if and only if $s = 1$, i.e. $q = 3$. $\varphi(l + 1) = l$. 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 $t \in \mathbb{N}$.