Reflexivity: For $R$ to be reflexive, we need $(f, f) \in R$ for all $f \in A$. This means $f(0) = f(1)$ and $f(1) = f(0)$, which is true for all $f$. Thus, $R$ is reflexive.

Symmetry: For $R$ to be symmetric, if $(f, g) \in R$, then $(g, f) \in R$. Given $(f, g) \in R$, we have $f(0) = g(1)$ and $f(1) = g(0)$. We need to check if $(g, f) \in R$, which means $g(0) = f(1)$ and $g(1) = f(0)$. Since $f(0) = g(1)$ and $f(1) = g(0)$, we have $g(1) = f(0)$ and $g(0) = f(1)$. Thus, $R$ is symmetric.

Transitivity: For $R$ to be transitive, if $(f, g) \in R$ and $(g, h) \in R$, then $(f, h) \in R$. Given $(f, g) \in R$, we have $f(0) = g(1)$ and $f(1) = g(0)$. Given $(g, h) \in R$, we have $g(0) = h(1)$ and $g(1) = h(0)$. We need to check if $(f, h) \in R$, which means $f(0) = h(1)$ and $f(1) = h(0)$. We have $f(0) = g(1) = h(0)$ and $f(1) = g(0) = h(1)$. Thus, $f(0) = h(0)$ and $f(1) = h(1)$. Therefore, $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation. However, the options do not include this. Let's re-examine the transitivity part.

We have $f(0) = g(1)$, $f(1) = g(0)$, $g(0) = h(1)$, and $g(1) = h(0)$. We want to show $f(0) = h(1)$ and $f(1) = h(0)$. From the given equations, $f(0) = g(1) = h(0)$ and $f(1) = g(0) = h(1)$. So, we have $f(0) = h(0)$ and $f(1) = h(1)$. This does NOT imply $f(0) = h(1)$ and $f(1) = h(0)$. Therefore, $R$ is NOT necessarily transitive.

Let's consider a counterexample for transitivity. Let $f(x) = x$, $g(x) = 1-x$, and $h(x) = x$. Then $f(0) = 0$, $f(1) = 1$, $g(0) = 1$, $g(1) = 0$, $h(0) = 0$, $h(1) = 1$. $(f, g) \in R$ since $f(0) = g(1) = 0$ and $f(1) = g(0) = 1$. $(g, h) \in R$ since $g(0) = h(1) = 1$ and $g(1) = h(0) = 0$. Now, we need to check if $(f, h) \in R$. We need $f(0) = h(1)$ and $f(1) = h(0)$. $f(0) = 0$ and $h(1) = 1$, so $f(0) \neq h(1)$. $f(1) = 1$ and $h(0) = 0$, so $f(1) \neq h(0)$. Thus, $(f, h) \notin R$, and $R$ is not transitive.

Since $R$ is reflexive and symmetric but not transitive, the correct option is (B) is incorrect and (D) is closest.

Let's re-examine reflexivity. For $R$ to be reflexive, we need $(f, f) \in R$ for all $f \in A$. This means $f(0) = f(1)$ and $f(1) = f(0)$. This is NOT true for all $f$. For example, if $f(x) = x$, then $f(0) = 0$ and $f(1) = 1$, so $f(0) \neq f(1)$. Thus, $R$ is not reflexive.

Since $R$ is symmetric but neither reflexive nor transitive, the correct option is (B).