RSA Encryption & Anonymity
I'll just expand one of the attacks mentioned in D.W.'s comment; this attack seems the most relevant to your question, but other attacks may exist according to the parameters you actually use.

Ideally, an encryption can be viewed as a permutation $\pi_i$ (determined by $e_i$), that maps $\{0,\ldots,n-1\}$ to itself. If you don't know the two permutations and you don't know the messages $m_i$, there is no way to tell if you used the permutation $\pi_1$ or $\pi_2$ by only seeing a small number of (image only) samples.

Under certain assumptions, RSA is a good encryption. This means you cannot decrypt a codeword $c$ and find $m$ such that $m^e=c \mod n$, even if you know $e$. Of course, information-theoretically, knowing $e$ gives you all the information you need to find $m$ from $c$–it fully specifies the permutation. But for a computationally-bounded Eve, we assume this is impossible.

However, if $n_1 \ne n_2$, Eve gets a certain advantage in distinguishing ecryptions in the two settings. By definition, $c_1 \in \{0,\ldots, n_1-1\}$ while $c_2 \in \{0, \ldots, n_2-1\}$. So if $n_1$ is very different from $n_2$ (say, $n_2&gt;n_1$), and Eve is lucky to see $c_i \in \{n_1, \ldots, n_2-1\}$, she knows that this message was encrypted using $(e_2,n_2)$. Assuming RSA is close to a random permutation, then this has probability of about $1/(n_2-n_1)$ per sample.