Let $$f:[1,\infty) \to [1,\infty)$$ be defined by $$f(x) = (x-1)^4 + 1$$. among the two statements:
(I) The Set $$S = \{x \in [1,\infty) : f(x) = f^{-1}(x)\}$$ contains  exactly two elements and 
(II) The Set $$S = \{x \in [1,\infty) : f(x) = f^{-1}(x+1)\}$$ is an empty set,

The function is $$f(x)=(x-1)^{4}+1$$ defined on $$[1,\infty)$$ with range $$[1,\infty)$$.
Derivative $$f'(x)=4(x-1)^{3}\ge 0$$ and $$f'(x)\gt 0$$ for $$x\gt 1$$, so $$f$$ is strictly increasing and hence bijective. Consequently the inverse exists and is

$$y=f(x)\;\Longrightarrow\;y-1=(x-1)^{4}\;\Longrightarrow\;x-1=(y-1)^{\frac14}$$
Therefore $$f^{-1}(y)=1+(y-1)^{\frac14}\qquad\bigl(y\in[1,\infty)\bigr).$$

Find $$S_1=\{x\in[1,\infty):f(x)=f^{-1}(x)\}.$$

Set $$f(x)=f^{-1}(x)\;\Longrightarrow\;(x-1)^{4}+1=1+(x-1)^{\frac14}$$ $$\Longrightarrow\;(x-1)^{4}=(x-1)^{\frac14}.$$

Put $$t=x-1\;(t\ge 0).$$ Then $$t^{4}=t^{\frac14}.$$ If $$t=0$$ the equality holds.
If $$t\gt 0$$ divide by $$t^{\frac14}$$ to get $$t^{\frac{15}{4}}=1\; \Longrightarrow\;t=1.$$ Thus $$t=0\;(x=1)$$ or $$t=1\;(x=2).$$ There are exactly two elements, $$\{1,2\}.$$
Hence statement (I) is TRUE.

Let $$S_2=\{x\in[1,\infty):f(x)=f^{-1}(x+1)\}.$$

The condition becomes$$(x-1)^{4}+1=1+\bigl((x+1)-1\bigr)^{\frac14}=1+x^{\frac14}$$ $$\Longrightarrow\;(x-1)^{4}=x^{\frac14},\qquad x\ge 1.$$ Define $$g(x)=(x-1)^{4}-x^{\frac14}.$$

Evaluate at two convenient points:
$$g(1)=(1-1)^{4}-1^{\frac14}=0-1=-1\lt 0,$$
$$g(3)=(3-1)^{4}-3^{\frac14}=2^{4}-3^{\frac14}=16-3^{0.25}\gt 16-2=14\gt 0.$$

Since $$g(x)$$ is continuous on $$[1,3]$$ and changes sign between $$x=1$$ and $$x=3$$, by the Intermediate Value Theorem there exists at least one $$c\in(1,3)$$ such that $$g(c)=0.$$ In other words, $$f(c)=f^{-1}(c+1)$$ for some $$c\in(1,3).$$
Thus $$S_2$$ is \emph{not} empty and statement (II) is FALSE.

Therefore, only statement (I) is true.
Option A which is: Only (I) is TRUE.