We show that (I) need not imply that $a = b = c$ whereas (II) always implies $a = b = c$.

Observe that $a(b^3 + c^3) = b(c^3 + a^3)$ gives $c^3(a - b) = ab(a^2 - b^2)$. This gives either $a = b$ or $ab(a + b) = c^3$. Similarly, $b = c$ or $bc(b + c) = a^3$. If $a \neq b$ and $b \neq c$, we obtain
$$ab(a + b) = c^3, \quad bc(b + c) = a^3.$$
Therefore
$$b(a^2 - c^2) + b^2(a - c) = c^3 - a^3.$$
This gives $(a-c)(a^2 + b^2 + c^2 + ab + bc + ca) = 0$. Since $a$, $b$, $c$ are positive, the only possibility is $a = c$. We have therefore 4 possibilities: $a = b = c$; $a \neq b$, $b \neq c$ and $c = a$; $b \neq c$, $c \neq a$ and $a = b$; $c \neq a$, $a \neq b$ and $b = c$.

Suppose $a = b$ and $b$, $a \neq c$. Then $b(c^3 + a^3) = c(a^3 + b^3)$ gives $ac^3 + a^4 = 2ca^3$. This implies that $a(a-c)(a^2 - ac - c^2) = 0$. Therefore $a^2 - ac - c^2 = 0$. Putting $a/c = x$, we get the quadratic equation $x^2 - x - 1 = 0$. Hence $x = (1 + \sqrt{5})/2$. Thus we get
$$a = b = \left( \frac{1 + \sqrt{5}}{2} \right) c, \quad c \text{ arbitrary positive real number.}$$
Similarly, we get other two cases:
$$b = c = \left( \frac{1 + \sqrt{5}}{2} \right) a, \quad a \text{ arbitrary positive real number;}$$
$$c = a = \left( \frac{1 + \sqrt{5}}{2} \right) b, \quad b \text{ arbitrary positive real number.}$$
And $a = b = c$ is the fourth possibility.

Consider (II): $a(a^3 + b^3) = b(b^3 + c^3) = c(c^3 + a^3)$. Suppose $a$, $b$, $c$ are mutually distinct. We may assume $a = \max\{a, b, c\}$. Hence $a > b$ and $a > c$. Using $a > b$, we get from the first relation that $a^3 + b^3 < b^3 + c^3$. Therefore $a^3 < c^3$ forcing $a < c$. This contradicts $a > c$. We conclude that $a$, $b$, $c$ cannot be mutually distinct. This means some two must be equal. If $a = b$, the equality of the first two expressions give $a^3 + b^3 = b^3 + c^3$ so that $a = c$. Similarly, we can show that $b = c$ implies $b = a$ and $c = a$ gives $c = b$.

Alternate for (II) by a contestant:

We can write
$$\begin{aligned} \frac{a^3}{c} + \frac{b^3}{c} &= \frac{c^3}{a} + a^2, \\ \frac{b^3}{a} + \frac{c^3}{a} &= \frac{a^3}{b} + b^2, \\ \frac{c^3}{b} + \frac{a^3}{b} &= \frac{b^3}{c} + c^2. \end{aligned}$$
Adding, we get
$$\frac{a^3}{c} + \frac{b^3}{a} + \frac{c^3}{b} = a^2 + b^2 + c^2.$$

Using C-S inequality, we have
$$(a^2 + b^2 + c^2)^2 = \left( \frac{\sqrt{a^3}}{\sqrt{c}} \cdot \sqrt{ac} + \frac{\sqrt{b^3}}{\sqrt{a}} \cdot \sqrt{ba} + \frac{\sqrt{c^3}}{\sqrt{b}} \cdot \sqrt{cb} \right)^2 \\ \le \left( \frac{a^3}{c} + \frac{b^3}{a} + \frac{c^3}{b} \right) (ac + ba + cb) \\ = (a^2 + b^2 + c^2)(ab + bc + ca).$$
Thus we obtain
$$a^2 + b^2 + c^2 \le ab + bc + ca.$$
However this implies $(a-b)^2 + (b-c)^2 + (c-a)^2 \le 0$ and hence $a = b = c$.