Is $\operatorname{Hom}_{RK}(R,RX)\cong R\otimes_{RK} RX$? Suppose a group $K$ acts on a finite set $X$, and let $RX$ be the permutation module of $X$ over a commutative unital ring $R$. Is it true that 
$$
\operatorname{Hom}_{RK}(R,RX)\cong R\otimes_{RK} RX?
$$
My motivation is that if this isomorphism is true I would be able to see that the invariants and coinvariants of a permutation module are isomorphic.

Yes, but not naturally. Both the invariants and coinvariants of a permutation module are free $R$-modules on the orbits of the action of $K$ on $X$. For the invariants, given an orbit $O$ the corresponding invariant is $\sum_{o \in O} o \in R[X]^K$, while for the coinvariants, given an orbit $O$ the corresponding coinvariant is any $o \in O$, standing in for its image in $R[X]_K$. 
On the other hand, there is a natural map $R[X]^K \to R[X]_K$ given as the composite
$$R[X]^K \to R[X] \to R[X]_K$$
but with respect to the two bases above it acts by multiplication by $|O|$ on the basis element corresponding to an orbit $O$, so it need not be an isomorphism in general, for example if $K = S_2, X$ is a two-element set, and $R = \mathbb{F}_2$. 
