Say Alice publishes $t$ unique signed messages with her private key $d$,
$$ m_1^d, m_2^d, m_3^d, ... , m_t^d \mod n $$
For $t = 2$, we have $m_1^d, m_2^d \mod n$ and if we let $x \equiv m_1m_2 \mod n$ then we can produce a valid signature,
$$ x^d \equiv (m_1m_2)^d \equiv m_1^d\times m_2^d \mod n $$
For $t$ signatures, I think we can have at most $\sum_{i=2}^{t}\binom{t}{i}$ new signatures. Is this correct?
This was inspired by a quiz question in which Alice published $t$ RSA signatures and we were asked to find a valid signature for $M \equiv m_1^{-1} \times m_2^{-1} \times m_3^{-1} \times ... \times m_t^{-1} \mod n$ and the solution was,
$$ M^d \equiv (m_1^{-1} \times m_2^{-1} \times m_3^{-1} \times ... \times m_t^{-1})^{d} \equiv (m_1^{d} \times m_2^{d} \times m_3^{d} \times ... \times m_t^{d})^{-1} \mod n $$
We assumed that all inverses existed. So, I was wondering why did this question involve inverses? Are they necessary to produce new signatures? Because as I showed above, I don't see why we could not do that otherwise.