1. Understand the partition of the set

   We are given $A=A_1\cup A_2\cup \cdots \cup A_k$ where the sets are pairwise disjoint: $A_i\cap A_j=\varnothing \quad \text{for } i\ne j.$

   So every element of $A$ belongs to exactly one of the sets $A_1,A_2,\dots,A_k$.

2. Interpret the relation

   The relation $R$ on $A$ is defined by $R=\{(x,y): y\in A_i \iff x\in A_i,\; 1\le i\le k\}.$

   This means that $(x,y)\in R$ exactly when $x$ and $y$ belong to the same block $A_i$.

   So we can rewrite it as: $xRy \iff x \text{ and } y \text{ lie in the same } A_i.$

3. Check reflexivity

   A relation is reflexive if $(x,x)\in R$ for every $x\in A$.

   Since every $x\in A$ lies in some unique $A_i$, clearly $x$ and $x$ belong to the same set $A_i$.

   Hence, $xRx \quad \forall x\in A.$

   Therefore, $R$ is reflexive.

4. Check symmetry

   A relation is symmetric if $xRy \implies yRx.$

   If $xRy$, then $x$ and $y$ are in the same set $A_i$. But then automatically $y$ and $x$ are also in the same set $A_i$.

   Hence, $xRy \implies yRx.$

   Therefore, $R$ is symmetric.

5. Check transitivity

   A relation is transitive if $xRy \text{ and } yRz \implies xRz.$

   Suppose $xRy$ and $yRz$.

   • $xRy$ means $x$ and $y$ are in the same $A_i$.
   • $yRz$ means $y$ and $z$ are in the same $A_j$.

   Since the sets $A_i$ are pairwise disjoint and $y$ belongs to both $A_i$ and $A_j$, we must have $A_i=A_j.$

   Therefore $x,y,z$ all belong to the same set, so $xRz$.

   Hence, $R$ is transitive.

6. Conclusion

   Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

7. Evaluate options

   • A: reflexive, symmetric but not transitive — false
   • B: reflexive, transitive but not symmetric — false
   • C: reflexive but not symmetric and transitive — false
   • D: an equivalence relation — true

Therefore, the correct option is $\boxed{D}.$