Let $A$ be a commutative ring, and $L, M, N$ be $A$-modules. Then is it true that
$$\text{Hom}\_A (L, M)\otimes\_A N \cong \text{Hom}\_A (L, M\otimes\_A N)$$
as $A$-modules? 


(Note that there is a natural morphism from the left to the right, I think it's not easy to check it is injective or surjective, but I didn't really do it; also, I think elements of the RHS are hard to decompose, so I don't hope for a (natural) arrow in the opposite direction.)


If this is not true, how about we assume that $A$ is a local ring and $N$ is a flat $A$-module or even a flat local $A$-algebra?


Could anyone give some hint or a proof, or a counterexample?


Other appropriate conditions that guarantee the isomorphism are appreciated.


$\textbf{Edit:}$ My main concern is the case when $A=\mathscr{O}\_{\mathbb{C}^n,0}=M, N=\mathscr{E}\_{\mathbb{C}^n,0}$, and $L$ is the stalk at $0\in \mathbb{C}^n$ of some coherent $\mathscr{O}\_{\mathbb{C}^n}$-module, where $\mathscr{O}\_{\mathbb{C}^n}$ and $\mathscr{E}\_{\mathbb{C}^n}$ mean the sheaves of holomorphic functions and complex-valued smooth functions on $\mathbb{C}^n$ respectively. The flatness of $\mathscr{E}\_{\mathbb{C}^n}$ over $\mathscr{O}\_{\mathbb{C}^n}$ can be found [here](https://books.google.com.hk/books?id=_o0l5D92wrcC&pg=PA211&lpg=PA211&dq=ring+of+smooth+functions+is+flat+over+the+ring+of+holomorphic+functions&source=bl&ots=-9kdf6v77-&sig=afXjcuINWBBsdTWm55qz8wKqSyk&hl=en&sa=X&ved=0ahUKEwiDurmXmeTJAhXGN6YKHTvWBj8Q6AEIKzAD#v=onepage&q=ring%20of%20smooth%20functions%20is%20flat%20over%20the%20ring%20of%20holomorphic%20functions&f=false) (Theorem 7.2.1), which cites Bernard Malgrange's book 'Ideals of differentiable functions' (page 88, Coro 1.12), and [here](https://mathoverflow.net/questions/226136/is-the-sheaf-of-smooth-functions-flat) is another discussion on MO.



Does module Hom commute with tensor product in the second variable?
The example $A=M=\mathbb{Z}$, $L=N=\mathbb{Q}$ shows that the answer is negative: we have $\text{Hom}(L,M)=0$ so $\text{Hom}\_A(L,M)\otimes\_AN=0$, but $\text{Hom}\_A(L,M\otimes\_AN)=\text{Hom}\_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q})=\mathbb{Q}\neq 0$


