Solution:
Since $\pi$ sends pairs $(m, n)$ with $m+n$ prime to pairs $(m', n')$ with $m'+n'$ prime, and there are only finitely many such pairs, we conclude that if $m+n$ is composite, then so is $\pi(m)+\pi(n)$. Also note that $2 \pi(1)=\pi(1)+\pi(1)$ is prime because $2=1+1$ is prime. Thus, $\pi(1)=1$.

Now, since $1+2, 1+4, 1+6$, and $1+10$ are all prime, we know that $\pi(2), \pi(4), \pi(6)$, and $\pi(10)$ are all even. Additionally, since $8+2, 8+6, 8+6$, and $8+10$ are all composite, it is not hard to see that $\pi(8)$ must also be even. Therefore $\pi$ preserves parity.

Now, draw a bipartite graph between the odd and even numbers where we have an edge between $a$ and $b$ if and only if $a+b$ composite. We now only need to compute automorphisms of this graph that fix $1$. Note that the edges are precisely $1-8-7-2, 3-6-9$, and $4-5-10$. Since $1$ is a fixed point of $\pi$, we know that $\pi$ fixes $1,8,7$, and $2$. Additionally, $\pi(6)=6$ and $\pi(5)=5$. We can swap $3$ and $9$, as well as $4$ and $10$. Thus, there are $2 \cdot 2=4$ possible permutations.