First we claim that $p : \mathcal{R}(f) \to \mathcal{R}(f)$ is a surjective function, where by $\mathcal{R}(f)$ we mean the range of the function $f$.
$$y \in \mathcal{R}(f) \Rightarrow \exists x_0 \in \mathbb{R}; f(x_0) = y \Rightarrow p(y) = p(f(x_0)) = f(p(x_0)) \Rightarrow p(y) \in \mathcal{R}(f)$$
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So $p: \mathcal{R}(f) \to \mathcal{R}(f)$ is indeed a function. Let $y \in \mathcal{R}(f)$ be an arbitrary element. Thus, there is some $x_0 \in \mathbb{R}$ such that $f(x_0) = y$. Note that since $p$ has odd degree there is $z \in \mathbb{R}$ such that $p(z) = x_0$ and therefore,
$$y = f(x_0) = f(p(z)) = p(f(z))$$
It means that $y$ is the image of $f(z) \in \mathcal{R}(f)$ under $p$ and so $p: \mathcal{R}(f) \to \mathcal{R}(f)$ is surjective.

a) Assume to the contrary that $\mathcal{R}(f)$ is not finite. Since $\deg(P)$ is odd we can find $N > 0$ such that
$$x > N \Rightarrow p(x) > x, \quad x < -N \Rightarrow p(x) < x$$
If $\mathcal{R}(f)$ contains infinitely many positive and infinitely many negative elements, we can choose $x_0, y_0 \in \mathcal{R}(f)$ such that $y_0 < -N < N < x_0$. Note that an element $b \in p^{-1}(a)$, where $a \in \mathcal{R}(f) \cap [y_0, x_0]$ must be in $\mathcal{R}(f) \cap [y_0, x_0]$. Because if $b < y_0$, then $p(b) = a < b < y_0$ and if $x_0 < b$, then $p(b) = a > b > x_0$ which are not possible. On the other hand, $f(x_0) > x_0$ and $f(y_0) < y_0$. Now since $\mathcal{R}(f) \cap [y_0, x_0]$ is a finite set (all its elements are integers), there must be some element which is not covered by elements of $\mathcal{R}(f)$ under $p$ and this contradicts because $p: \mathcal{R}(f) \to \mathcal{R}(f)$ was surjective. The argument in the case which $\mathcal{R}(f)$ contains only finitely many positive or negative integers are similar.

b) Since $\deg(p(x)) > 1$ is an odd number, $p(x) - x$ has odd degree. Therefore, it has at least one real root $x_0$. Suppose that this polynomial has no real root other than $x_0$. Thus, $p(x) > x$ for $x > x_0$ and $p(x) < x$ for $x < x_0$.
On the other hand, we have $p(f(x_0)) = f(p(x_0)) = f(x_0)$ and since $x_0$ was the unique solution of $p(x_0) = x_0$, we get $f(x_0) = x_0$. Suppose that $\mathcal{R}(f) = \{x_{-m} < \cdots < x_{-1} < x_0 < x_1 < \cdots < x_n\}$. If $n > 0$, $p(x_n) > x_n$ and this contradicts because $p(\mathcal{R}(f)) \subseteq \mathcal{R}(f)$. So $n = 0$. Similarly, $m = 0$ and therefore, $|\mathcal{R}(f)| = 1$. But we assumed that $|\mathcal{R}(f)| \ge 2$.

c) Suppose that $z, y_1, \dots, y_m$ are $n = m+1$ arbitrary distinct integers. Using Lagrange Interpolation, we can find a polynomial $p(x) \in \mathbb{R}[x]$ such that
$$p(z) = z, \quad p(y_1) = y_2, \quad p(y_2) = y_3, \quad \dots, \quad p(y_m) = y_1$$
If it is necessary by changing $x^k(x-z)(x-y_1)\cdots(x-y_m) + p(x)$ with $p(x)$ ($k \in \mathbb{N}$), we can assume that $p$ is a monic polynomial of odd degree. Now we define function $f$ as follows. If there is some integer $k \ge 0$ and $1 \le i \le m$ such that $p^k(x) = y_i$, set $f(x) = y_{i-k}$ (Indices are assumed to be modulo $m$). Note that if $p^{k_1}(x) = y_i$ and $p^{k_2}(x) = y_j$, $y_{i-k_1} = y_{j-k_2}$. Thus, there is no ambiguity in the definition. Otherwise, define $f(x) = z$. We claim that this definition satisfies problem statement.
* If $f(x) = y_i$, there is some integer $k$ such that $p^k(x) = y_{i+k}$. Now $p^k(p(x)) = p(y_{i+k}) = y_{i+k+1}$ and so $f(p(x)) = y_{i+k+1-k} = y_{i+1} = p(y_i) = p(f(x))$.
* If $f(x) = z$, then obviously $f(p(x)) = z$ and so $f(p(x)) = z = p(z) = p(f(x))$.
Therefore, the proof is complete.