In other words, find the number of such injective polynomial functions $f\colon \mathbb{Z}_{2013} \to \mathbb{Z}_{2013}$ (which therefore will be bijections).

I think the answer is $3600$. Here is a solution: Let $(a, b)$ be called a nice pair if $f(x)=ax^3+bx$ satisfies the conditions stated in the problem. Firstly, we shall prove a lemma: If $(a, b)$ is a nice pair then $\gcd (671, a(x^2+xy+y^2)+b)=1$ for all integers $x$ and $y$. Proof: Let $p=11$ or $61$ and $p$ divides $a(x^2+xy+y^2)+b$ for some integers $x=u$ and $y=v$. Then we can find another pair $(x_0, y_0)$ such that $x_0 \not\equiv y_0 \pmod p$ and $u^2+uv+v^2 \equiv x_0^2+x_0y_0+y_0^2 \pmod p$. (The proof is easy) Then consider the system $x \equiv y \pmod {\frac{2013}{p}}$ and $x \equiv x_0 \pmod p, \quad y \equiv y_0 \pmod p$. By the Chinese Remainder Theorem this system has a solution $x=x_1$ and $y=y_1$ such that both $x_1$ and $y_1$ is an element of $\{1, 2, \ldots ,2013\}$ which means that $(a, b)$ is not a nice pair because $f(x_1)-f(y_1)=(x_1-y_1)[a(x_1^2+x_1y_1+y_1^2)+b]$ is divisible by $2013$. Now, we shall show that if $p=61$ or $p=11$ then $x^2+xy+y^2 \equiv c \pmod p$ has a solution for all integers $c$. For the proof assume there exists an integer $c$ such that $x^2+xy+y^2 \equiv c \pmod p$ has no solution in integers. Then it is easy to see that $x^2+xy+y^2 \equiv k^2c \pmod p$ has also no solution for all integers $k$ which are not divisible by $p$. Now take $k=1, \ldots ,\frac{p-1}{2}$ and consider the numbers $k^2c$. For all of these numbers, we have no solution and there are $\frac{p-1}{2}$ such numbers. Since all square residues have solutions (take $y=0$) and there are $\frac{p+1}{2}$ square residues modulo $p$ including zero, this means that for all nonsquare residues $c$, the equation must have no solution. However, for $p=11$ the number $10$ is not a square residue but $4^2+4+1=21 \equiv 10 \pmod {11}$ so for $c=10$ we have a solution which is a contradiction. For $p=61$ the number $2$ is not a square residue but $x^2+2x+4 \equiv 2 \pmod {61}$ has a solution since $-1$ is a square residue. Hence we again obtain a contradiction so the second lemma is also proved. Now, since $x^2+xy+y^2$ can take every value modulo $11$ and $61$ we must have $11 \mid a$ and $61 \mid a$ and also we must have $11 \nmid b$ and $61 \nmid b$ if $(a, b)$ is a nice pair. So, $a$ can take only three values. We will consider each case separately: If $a=671$ then we must have $3 \mid 671(x^2+xy+y^2)+b \: \Longrightarrow \: 3 \mid x-y$ which means $3 \nmid b-1$ If $a=2 \cdot 671$ then we must have $3 \mid 2 \cdot 671(x^2+xy+y^2)+b \: \Longrightarrow \: 3 \mid x-y$ which means $3 \nmid b-2$ If $a=3 \cdot 671$ then we must have $3 \mid 3 \cdot 671(x^2+xy+y^2)+b \: \Longrightarrow \: 3 \mid x-y$ which means $3 \nmid b$ If we count these possibilities, if i am not wrong, we get $2[2013-(61 \cdot 3 + 11 \cdot 3 -3)] = 3600$.

First, remark that it suffices for $f$ to be injective modulo $3,11$ and $61$. For modulo $3$ simply note that we require $a + b \not \equiv 0 \pmod{3}$. For modulo $11$ or $61$ we require for some $m \neq n$ that \[\begin{eqnarray*} am^3 + bm & \equiv & an^2 + bn \pmod{p} \\ a(m-n)(m^2 + mn + n^2) & \equiv & -b(m-n) \pmod{p} \\ a(m^2 + mn + n^2) & \equiv & -b \pmod{p}\] Now, what values can $m^2 + mn + n^2$ take modulo $p$? It is a simple exercise to show all values modulo $p$ can be obtained (just express it as $\left ( m + \frac{n}{2} \right )^2 + 3 \left ( \frac{n}{2} \right )^2$ and then as $11,61 \equiv \pm 1 \pmod{12}$ we can transform the problem to what values modulo $p$ can be expressed as $m^2 + n^2$, which is known to be all of them). Therefore the only way this problems works out is if $a \equiv 0 \pmod{p}$ and $b \not \equiv 0 \pmod{p}$. Therefore we simply require that $671|a$ and $3 \nmid (a+b)$. For each value of $a$ there are $2$ values of $a$ modulo $3$, 10 modulo $11$ and $60$ modulo $61$ so the answer should simply be $3 \cdot 2 \cdot 10 \cdot 60 = 3600$. Note: To show $m^2 + mn + n^2$ takes all values modulo $p$ without relying on $11,61 \equiv \pm 1 \pmod{12}$ is not hard, but I'm lazy and felt like reducing it to an already solved problem.

dinoboy wrote: Note: To show $m^2 + mn + n^2$ takes all values modulo $p$ without relying on $11,61 \equiv \pm 1 \pmod{12}$ is not hard, but I'm lazy and felt like reducing it to an already solved problem. I think for all prime numbers different than $3$ it is true.

Yeah it is true for all primes (hence the "without relying on" part), but I was too lazy to put the proof.

$\left( {\frac{{ - 3}}{{11}}} \right) = - 1$，so ${c^2} + cd + {d^2} \equiv 0(\bmod 11) \Rightarrow c \equiv d \equiv 0(\bmod 11)$，hence $a\not \equiv 0(\bmod 11),b \equiv 0(\bmod 11)$ satisfies the problem. So the answer should be $3 \cdot 2 \cdot 20 \cdot 60 = 7200$.

Oops, yunxiu is correct. I completely forgot about the condition that $m,n$ need to be distinct when deducing which numbers are expressible as $m^2 + mn + n^2$.

Claim. $a$ is divisble by $61$ Proof. We first claim that there exists a solution $x\neq y$ to $$a(x^2+xy+y^2)\equiv -b\pmod{61}$$Indeed, since $a\neq 0$ we can assume it is equal to $1$. Fix $y=2c$ and the equation is equivalent to $$(x+c)^2\equiv -b-3c^2$$Therefore, since there are $31$ quadratic residues mod 61, some of $-b-3c^2$ must be quadratic residue, so there is a solution (if $b=3c^2$ we check that $b$ is a quadratic residue) Now pick $(m,n)\equiv (x,y)\pmod 61$ and $33|m-n$, then $2013\nmid m-n$ but $$(m-n)(m^2+mn+n^2+b)\equiv 0\pmod{2013}$$contradiction. $\blacksquare$ Similarly we can show that if $11\nmid b$ then $11|a$. We divide into two cases: Case I: $11|a$ Then $(b,671)=1$, so if $2013|f(m)-f(n)$ we have $$2013|(m-n)(a(m^2+mn+n^2)+b)$$Hence $671|m-n$, and $3\nmid m-n$. Therefore, $m^2+mn+n^2\equiv 1\pmod 3$, $f$ satisfy the condition if and only if $3\nmid a+b$. For each $a$ there are $\varphi(2013)$ choices, so totally there are $$3\times \varphi(2013)$$choices. Case II: $11|b$ Then $(11,a)=1$. Notice that $11\nmid x^2+xy+y^2+b$, hence $$2013|(m-n)(a(m^2+mn+n^2)+b)$$implies $671|m-n$ and $3\nmid a+b$. For each $a$, there are $$2013\cdot\frac{1}{11}\cdot\frac{60}{61}\cdot\frac{2}{3}=120$$coices for $b$, there are $$2013\cdot\frac{1}{61}\cdot\frac{10}{11}=30$$choices, so the answer is $$30\cdot 120+3\varphi(2013)=7200$$