When multiplying by $i$ in inequalities, does the sign flip?

Take $1>-1$, when you multiply by $-1$, to get $-1<1$ you have to flip the sign.

What if you are multiplying by $i$? Would you flip the signs? If you didn't it would net the expression $i>-i$. If you continued that rationale and not flip the signs a second time, you could come to the conclusion that $-1>1$, a clear contradiction. If you were to flip the signs, then from $1>-1$ you would get $i<-i$. Again, if you continue this again, it comes up with the contradiction of $-1>1$. Is it just an undefined inequality? I have no idea.

Is it just an undefined inequality

Yes, that's it. We do not define inequalities (i.e., order relation) between complex numbers. There is no way to define them in a way consistent with arithmetic operations. For example, if we decided that "$i>0$" or "$i<0$", multiplying both sides by $i$ would give $-1>0$ in either case.

And if we accepted $-1>0$ as true, it would lead deeper down the hole. For one thing, it would imply $(-1)(-1)>0$, hence $1>0$. If both $1>0$ and $-1>0$, we don't have consistency with arithmetics anymore. Adding the inequalities would give $0>0$, and so forth.

Well, as far as I understand, you can't really say that one complex number is 'bigger' than another. You could compare the moduli with inequalities, but not the complex numbers themselves.

e.g. $$|1+i|>1$$

Hope I've helped!