Since $5$ is a divisor of $2015$, we have for any real $z$:
$$\begin{aligned} f^{25}(z) &= f^5(f^5(f^5(f^5(z)))) = 2015 - f^5(f^5(f^5(f^5(z)))) \\ &= 2015 - (2015 - f^5(f^5(f^5(z)))) = f^5(f^5(f^5(z))) = \dots = f^5(z) = 2015 - z. \end{aligned}$$
Since $13$ is also a divisor of $2015$, we have for any real $z$:
$$f^{26}(z) = f^{13}(f^{13}(z)) = 2015 - f^{13}(z) = 2015 - (2015 - z) = z.$$
Consequently $z = f^{26}(z) = f(f^{25}(z)) = f(2015 - z)$. Any real number $x$ can be written as $2015 - z$ for $z = 2015 - x$. Hence $z = f(2015 - z)$ implies $f(x) = 2015 - x$ for any real $x$.

Finally check that the function $f(x) = 2015 - x$ satisfies the conditions of the problem. Let $d$ be a divisor of $2015$ greater than $1$. Then $d$ is odd, i.e. $d = 2c + 1$ for a positive integer $c$. Since $f^2(x) = 2015 - (2015 - x) = x$, we have $f^{2c}(x) = f^2(f^2(\dots f^2(x)\dots)) = x$, which implies $f^d(x) = f(f^{2c}(x)) = f(x) = 2015 - x$.