Proof of ${\rm Hom}_R(R,R)=R$ If $R$ is commutative then $${\rm Hom}_R(R,R)=R$$
Proof : Let $A$ be an index set s.t. $$
R = \oplus_{a\in A} R_a $$ where $R_a$ is commutative
Hence  $$ {\rm Hom}_R(R,R) = \prod_{a\in A} {\rm Hom}_R (R_a,R) $$
For $f \in {\rm Hom}_R (R_a,R)$, if 
$f(x)=(y_b),\ y_b\in R_b$ so $$f(x^2)=x(y_b)=xy_a $$
I do not prove this
[add] In problem, $R$ is commutative ring with $1$. 
So for $r\in R$, $$f_r(x)=rx,\ f_r\in {\rm Hom}_R(R,R)$$    
Define $F : R\rightarrow {\rm Hom}_R(R,R),\ F(r)=f_r$
Then $F(r+s)=f_{r+s} = f_r+f_s,\ F(rs)=f_{rs}=f_rf_s$ and if $f_r=f_s$ then $$ f_r(1)=f_s(1)\Rightarrow r=s $$
 A: Commutativity does not matter, although a little attention to the side and identity matter.
The right module $R_R$ is a free module on the basis $1$. Thus $1$ every function $f:\{1\}\to R$ uniquely determines an $R$-homomorphism $f:R\to R$.
Let's discuss it in a simpler way, if you do not understand free modules very well.
Notice that given an element $a\in R$, the map $\ell_a:R\to R$ given by $\ell_a(r)=ar$ is a right module homomorphism.
Given a homomorphism $\phi:R\to R$, we can see that $\phi(1)=a$, and then by linearity $\phi(r)=\phi(1)r=ar=\ell_a(r)$ for all $r$. So in summary, left multiplication is a right module homomorphism, and every homomorphism from $R$ to $R$ is of that form.
So the logical map to use is $\theta:R\to End(R_R)$ given by $\theta(a)=\ell_a$. I leave it to you to check that this map is additive, multiplicative, and an isomorphism.

Warning: if you used the left module $_RR$ instead, then you should carefully show that $End(_RR)\cong R^{op}$, the opposite ring of $R$ rather than $R$ itself! Of course, if you want to stick to commutative rings this does not matter at all.
