Integrality and closed form for a greatest-common-divisor exponential sum

As stated, the conjecture is false because no restriction is imposed on $k$. Take

$$n=3,\qquad a=2,\qquad d=1,\qquad k=3.$$

Then $\varphi(n)=2$ and $M_1=M_2=1$, so

$$\frac1{\varphi(n)} \sum_{i=1}^{\varphi(n)}\gcd(n,M_i)\zeta_3^i =\frac{\zeta_3+\zeta_3^2}{2} =-\frac12\notin\mathbb Z.$$

The natural missing hypothesis in the preceding cyclotomic-product formula is $k\mid\varphi(n)$. With this correction, the conjecture holds and admits an explicit closed form.

Write $L=\varphi(n)$, put

$$P_d(X)=1+X+\cdots+X^{d-1},\qquad M_i=P_d(a^i),$$

and for every $s\mid n$ with $\gcd(s,a)=1$, let $h_s=\operatorname{ord}_s(a)$, with $h_1=1$. Let

$$c_t(m)=\sum_{\substack{1\leq u\leq t\\gcd(u,t)=1}} e^{2\pi imu/t}$$

denote the Ramanujan sum. For every $k\mid L$,

$$\boxed{ \frac1L\sum_{i=1}^L\gcd(n,M_i)\zeta_k^i = \sum_{\substack{s\mid n,\ \gcd(s,a)=1\k\mid h_s}} \frac{\varphi(s)}{h_s} \sum_{\substack{t\mid h_s\s\mid P_d(a^{h_s/t})}} c_t(h_s/k). } \tag{1}$$

To prove this, first note that if $h=\operatorname{ord}_s(a)$, $\gcd(u,h)=1$, and $g=a^r$, then

$$s\mid P_d(g)\quad\Longleftrightarrow\quad s\mid P_d(g^u). \tag{2}$$

Work modulo each $p^e\Vert s$, using

$$P_d(g^u)P_u(g)=P_u(g^d)P_d(g). \tag{3}$$

If $g\equiv1\pmod{p^e}$, both divisibility conditions reduce to $p^e\mid d$. If $g\equiv1\pmod p$ but $g\not\equiv1\pmod{p^e}$, the order of $g$ modulo $p^e$ is a nontrivial $p$-power, so $p\mid h$ and $p\nmid u$. Thus both $P_u(g)$ and $P_u(g^d)$ are units modulo $p$, and (3) proves (2).

Finally, if $g\not\equiv1\pmod p$, then $P_u(g)$ is a unit modulo $p$, since otherwise $g^u\equiv1\pmod p$ contradicts $\gcd(u,h)=1$. The same holds for $P_u(g^d)$, except possibly when $g^d\equiv1\pmod p$ and $p\mid u$. In this exceptional case $p\nmid h$, and reduction modulo $p$ is injective on the cyclic subgroup generated by $a$ modulo $p^e$, because its kernel is a $p$-group. Therefore $g^d\equiv1\pmod{p^e}$; since $g-1$ and $g^u-1$ are units, both $P_d(g)$ and $P_d(g^u)$ vanish modulo $p^e$. This establishes (2).

The reduction map

$$(\mathbb Z/h\mathbb Z)^\times \longrightarrow (\mathbb Z/t\mathbb Z)^\times$$

is surjective for every $t\mid h$. Consequently (2) shows that $s\mid P_d(a^r)$ depends only on the order $t$ of $a^r$, equivalently on $\gcd(r,h)$.

Now expand

$$\gcd(n,M_i)=\sum_{\substack{s\mid n\s\mid M_i}}\varphi(s).$$

Any $s$ sharing a prime factor with $a$ never contributes. For the remaining $s$, Euler's theorem gives $h_s\mid\varphi(s)\mid L$. Grouping indices modulo $h_s$, the contribution vanishes unless $k\mid h_s$; otherwise it equals

$$\frac{\varphi(s)}{h_s} \sum_{r=1}^{h_s} \mathbf1_{s\mid P_d(a^r)}\zeta_k^r.$$

The residues giving elements of order $t\mid h_s$ are

$$r=\frac{h_s}{t}u,\qquad\gcd(u,t)=1.$$

They all contribute exactly when $s\mid P_d(a^{h_s/t})$, and their exponential sum is $c_t(h_s/k)$. This proves (1). Since $h_s\mid\varphi(s)$ and every Ramanujan sum is integral, the right-hand side belongs to $\mathbb Z$.

Thus the unrestricted statement has the explicit counterexample $-1/2$, whereas the intended restricted statement $k\mid\varphi(n)$ is true with the closed form (1).