More than one modular multiplicative inverse possible?

Are you sure that $65$ and $-12$ are different?

Think about how integers "name" integers mod $k$: every integer $n$ corresponds to the set $\{m: m=n$ (mod $k)\}$ of integers which are equivalent to it mod $k$. Two different integers can correspond to the same set; for example, $1$ and $3$ represent the same thing in mod $2$.

The right way to think about this, in my opinion, is the following. "The integers mod $k$" is really a structure in its own right, usually denoted "$\mathbb{Z}/k\mathbb{Z}$" - elements of this structure aren't integers, but rather sets of integers. For instance, $\mathbb{Z}/2\mathbb{Z}$ (integers mod $2$) has two elements, "even numbers" and "odd numbers."

Specifically, elements of $\mathbb{Z}/k\mathbb{Z}$ are equivalence classes. When we say "$x=y$ (mod $k)$," what we really mean is that the equivalence class "$x$ (mod $k$)" is the same as the equivalence class "$y$ (mod $k$)," and the point in this particular question is that $65=-12$ (mod $77)$.

$$-12 \equiv 65 \pmod{77}$$

To see that notice that $65-(-12)=77$.

The multiplicative inverse should be a positive integer. You are correct that $$-12 * 32 \pmod{77} = 1.$$ But $65 * 32 \pmod{77} = 1$ as well. If you come up with a negative multiplicative inverse, then you need to take

$$\text{negative multiplicative inverse}\pmod{77} \,=\,\, \text{ positive multiplicative inverse}.$$

By all means verify this for yourself. Test that both function as multiplicative inverses. But for the test, only the positive one is correct.