Let $\prod_{i=1}^{m} x_i$ denote the product of $m$ real numbers $x_1, x_2, \dots, x_m$. For any positive integer $N$, let $\text{ord}_2 N$ denote the maximum nonnegative integer $\ell$ such that $2^\ell$ divides $N$.

$n=1$ does not meet the assumption since $\frac{1^{\phi(1)} - 1}{d(1)^5} = 0$ is an integer. In the following we assume $n \ge 2$. If distinct prime numbers $p_1, p_2, \dots, p_k$ and positive integers $e_1, e_2, \dots, e_k$ satisfy $n = \prod_{i=1}^{k} p_i^{e_i}$, then $\phi(n) = \prod_{i=1}^{k} p_i^{e_i-1}(p_i - 1)$.

When $n$ is even, the assumption that $\frac{\phi(n)^{d(n)} + 1}{n}$ is an integer implies that $\phi(n)$ is odd, thus $n=2$. On the other hand, $n=2$ satisfies the assumptions since $\frac{\phi(2)^{d(2)} + 1}{2} = \frac{1^2+1}{2} = 1$ and $\frac{2^{\phi(2)} - 1}{d(2)^5} = \frac{2^1 - 1}{2^5} = \frac{1}{32}$.

In the following we assume $n$ is odd and $n \ge 3$. The assumption that $\frac{\phi(n)^{d(n)} + 1}{n}$ is an integer implies that $n$ and $\phi(n)$ are coprime, thus $n$ is square-free. Then $n = \prod_{i=1}^{k} p_i$ with distinct odd primes $p_1, \dots, p_k$, thus $d(n) = 2^k$. Now we need the following lemma.

Lemma. For any odd integer $x \ge 3$ and any positive integer $y$, there holds $\text{ord}_2(x^y - 1) \ge \text{ord}_2(x-1) + \text{ord}_2 y$.

Proof. Suppose $y = 2^v \cdot s$ with a nonnegative integer $v$ and a positive odd integer $s$. Then $x^y - 1 = (x^s - 1) \prod_{i=0}^{v-1} (x^{2^i \cdot s} + 1)$. Since $x^s - 1 = (x-1)(x^{s-1} + \dots + x+1)$, $\text{ord}_2(x^s - 1) \ge \text{ord}_2(x-1)$.
Also, $\text{ord}_2(x^{2^i \cdot s} + 1) \ge 1$ for any $i$ since $x$ is odd, thus we obtain $\text{ord}_2 \left( \prod_{i=0}^{v-1} (x^{2^i \cdot s} + 1) \right) \ge v$.
Hence $\text{ord}_2(x^y - 1) \ge \text{ord}_2(x-1) + \text{ord}_2 y$. $\blacksquare$

The condition that $\frac{n^{\phi(n)} - 1}{d(n)^5} = \frac{n^{\phi(n)} - 1}{2^{5k}}$ is not an integer is equivalent to $\text{ord}_2(n^{\phi(n)} - 1) < 5k$. The above lemma implies that $\text{ord}_2(n^{\phi(n)} - 1) \ge \text{ord}_2(n-1) + \text{ord}_2(\phi(n))$. Since $\frac{\phi(n)^{d(n)} + 1}{n}$ is an integer, for any $p_i$ there holds $\phi(n)^{2k} \equiv -1 \pmod{p_i}$, thus $\phi(n)^{2^{k+1}} \equiv 1 \pmod{p_i}$. Let $t_i$ be the minimum positive integer $t$ such that $\phi(n)^t \equiv 1 \pmod{p_i}$. If a positive integer $\ell$ satisfies $\phi(n)^\ell \equiv 1 \pmod{p_i}$, then $t_i$ divides $\ell$; this is because when $r$ denotes the remainder of $\ell$ modulo $t_i$, then $\phi(n)^\ell \equiv \phi(n)^r \equiv 1 \pmod{p_i}$, which implies $r=0$ by the minimality of $t_i$. Now $t_i$ divides $2^{k+1}$ since $\phi(n)^{2^{k+1}} \equiv 1 \pmod{p_i}$. On the other hand $t_i$ does not divide $2^k$ since $\phi(n)^{2^k} \equiv -1 \not\equiv 1 \pmod{p_i}$. Hence we obtain $t_i = 2^{k+1}$.

By the Fermat's little theorem $\phi(n)^{p_i-1} \equiv 1 \pmod{p_i}$, thus $2^{k+1}$ divides $p_i - 1$. Hence $\text{ord}_2(\phi(n)) = \sum_{i=1}^{k} \text{ord}_2(p_i - 1) \ge k(k+1)$. On the other hand, by $n-1 = \prod_{i=1}^{k} p_i - 1 \equiv \prod_{i=1}^{k} 1 - 1 \equiv 0 \pmod{2^{k+1}}$ we obtain $\text{ord}_2(n-1) \ge k+1$. Therefore $5k > \text{ord}_2(n^{\phi(n)} - 1) \ge \text{ord}_2(n-1) + \text{ord}_2(\phi(n)) \ge (k+1)^2$. Then we obtain $k=1, 2$.

When $k=1$ we obtain $\phi(n) = n-1$ and $d(n) = 2$, thus $\frac{\phi(n)^{d(n)} + 1}{n} = \frac{(n-1)^2 + 1}{n} = n-2 + \frac{2}{n}$ is not an integer, which contradicts the assumption. When $k=2$, the condition $3+2 \cdot 3 \le \text{ord}_2(n-1) + \text{ord}_2(\phi(n)) < 5 \cdot 2$ implies that $\text{ord}_2(n-1) = 3$ and $\text{ord}_2(\phi(n)) = 6$. Since $p_1-1$ and $p_2-1$ are both multiples of $2^{2+1}$, $\text{ord}_2((p_1-1)(p_2-1)) = \text{ord}_2(\phi(n)) = 6$ implies $\text{ord}_2(p_1-1) = \text{ord}_2(p_2-1) = 3$. Thus $p_1 \equiv p_2 \equiv 9 \pmod{16}$. Hence $n-1 = p_1p_2-1 \equiv 0$ (mod 16), which contradicts $ord_2(n-1) = 3$. Thus there does not exist an odd integer $n \ge 3$ which satisfies the required conditions.

Hence the answer is $n = 2$.