Can an RSA private key have several public keys?
@pg1989 has not only the right answer but the only answer if you assume the goal is to have multiple RSA encryption exponents that correspond to the same decryption exponent.

\begin{align}
e_1 \cdot d &amp; \equiv 1 \pmod{\phi(N)}\\
e_2 \cdot d &amp; \equiv 1 \pmod{\phi(N)}\\
(e_1 - e_2) \cdot d &amp; \equiv 0 \pmod{\phi(N)}
\end{align}

But since $d$ is relatively prime to $\phi(N)$, by Euclid's lemma $\phi(N)$ must divide $e_1 - e_2$ :

$$e_1 \equiv e_2 \pmod{\phi(N)}$$

In other words, any two public RSA exponents that share a private exponent must differ by some multiple of $\phi(N)$.

I highly suspect this is what your professor was thinking. Although, as @Thomas points out if the multiple is small and someone gets hold of both public keys, they can factor $N$. Now, maybe if you use a big multiple...

[Update]

As @fgrieu notes in a comment, I have conflated $\phi(N) = (p-1)(q-1)$ with $\operatorname{LCM}(p-1, q-1)$. Replace the former with the latter in the above derivation for the right answer.