$\textbf{Answer}$

No, it is impossible.

$\textbf{Analysis}$

This looks like a two-state problem. Meanwhile, because there is an infinite number of ways to turn these coins, it is also likely to be an invariant problem. As a result, we will try to use a two-state model and look for an invariant.

Let number $1$ represent the status of heads up and $(-1)$ represent the status of heads down. Then, initially, we have nine $1$s. Their product (multiplying them together) is $1$. Flipping a coin will make its state transform from $1$ to $(-1)$, or from $(-1)$ to $1$. Either way, it can be modeled as multiplying $(-1)$ on the pre-state to get the post-state. For example, if a coin is heads up (i.e. state $1$), flipping it is equivalent to $1\times (-1)=-1$ which means its new state will be heads down $(-1)$.

Each operation involves flipping two coins. This means applying "multiplying by $(-1)$" twice. Because $(-1)\times(-1)=1$, therefore it will not alter their product. This means that the product of all the nine coins' status is always $1$. However, nine heads down will mean their product is $(-1)$. Hence, it is impossible.