Can one construct OTPs without using XOR? The typical version of the one-time-pad (OTP) uses XOR to combine a key pad and a message. ($c=m\oplus k$)

Now let's assume some other scenarios which have the practical application of blinding. Do the below "OTP" constructions provide perfect secrecy?


The key $r\in (0;n)$ is chosen uniformly at random such that $r^{-1}\bmod n$ exists and shared between both parties. Furthermore both parties share some common modulus $n$ which may be an RSA-modulus or a prime. For a message $m\in (0;n)$, is the construction $c=m\cdot r \bmod n$ perfectly secure?
$r\in (0;n)$ is chosen uniformly at random  and shared between both parties. Furthermore both parties share some common modulus $n$ which may be an RSA-modulus or a prime. For a message $m\in (0;n)$, is the construction $c=m + r \bmod n$ perfectly secure? (I strongly suspect the answer to be "yes" here as this is a "commonly" cited method for secret message transmission via radio where $n=10^4$)
Both constructions are not perfectly secure!



In an attempt to express things a bit more mathematically, I'd say you can implement a one-time-pad with a message $m$ and a key $r$, when


$m$ and $r$ are elements of $\mathbb{Z}/n\mathbb{Z}$, the additive group of integers modulo n, or when
$m$ and $r$ are elements of a group $G$ isomorphic to $\mathbb{Z}/n\mathbb{Z}$.


Such groups would be a natural choice to implement an OTP. 

There are however also other possibilities. For instance one can easily imagine a perfectly secure OTP construction where the key $r$ is an element of a group $G$ and the message $m$ is an element of a subgroup of $G$. (The key is then larger than the message).

Now, lets come back to your constructions. In both constructions you assume $r,m\in \{1,...,n-1\}$ using the unconventional notation $r,m\in(0;n)$.


Construction 1: You compute $c=r\cdot m $ mod $n$. If $n$ is a prime, then you have the multiplicative group of a prime field and it will work. As you already found out, it will not work when $n$ is composite.
Construction 2: You compute $c=r+ m $ mod $n$. It would be perfectly secure if you had $\mathbb{Z}/n\mathbb{Z}$. However, since you exclude the 0 element, you know that you always have $m\ne c$, and it is thus not perfectly secure. :)