For $R$-modules M, is $M\cong R^{\oplus n}\otimes_RM\cong M^{\oplus n}$? I wanted to explicitly give a bilinear map $R^{\oplus n}\times M\longrightarrow M^{\oplus n}$ when trying to prove that $R^{\oplus n}\otimes_RM\cong M^{\oplus n}$ and ended up with $(r_1,...,r_n,m)\mapsto ((\prod_{i=1}^nr_i)m,...,(\prod_{i=1}^nr_i)m)$ which seems to work in showing that every bilinear map to some $R$-module P : $R^{\oplus n}\times M\longrightarrow P$ factors through $M^{\oplus n}$. 
But I think the same works for $R^{\oplus n}\times M\longrightarrow M$, $(r_1,...,r_n,m)\mapsto (\prod_{i=1}^nr_i)m$. So is this true or did I do a mistake?
I'm in a commutative ring $R$ with identity. 
Edit: I probably should have made clear that I am trying to find a map that together with $M^{\oplus n}$ satisfies the universal property for the tensor product, and am not trying to give an isomorphism.

The map you wrote is not an isomorphism. It is instead a map that arguably parametrizes (though not uniquely) all multiples of the diagonal map $M\rightarrow M^{\oplus n}$ (where the multiple, depending on an element $(r_1,\ldots,r_n)$ of $R^{\oplus n}$, is $\prod_ir_i$). This is why, as you observe, you could make a similar argument for $R^{\oplus n}\times M \rightarrow M$.
Instead, what you want is the map $(r_1,\ldots,r_n,m) \mapsto (r_1m,\ldots,r_nm)$. You can check that this induces the isomorphism $R^{\oplus n}\otimes M \rightarrow M^{\oplus n}$ that you want. If you know about the distributive property of the tensor product, you can also deduce this same result by noting the following:
$$
\begin{array}{rcl}
R^{\oplus n}\otimes M & = & (R \oplus \cdots \oplus R)\otimes M\\
& \cong & (R\otimes M)\oplus \cdots \oplus (R\otimes M)\\
& \cong & M\oplus \cdots \oplus M\\
& = & M^{\oplus n}
\end{array}
$$
