Let $\mathbb X$ be the set of all bijective functions from the set $S=\{1,2,\cdots, n\}$ to itself. For each $f\in \mathbb X,$ define
\[T_f(j)=\left\{\begin{aligned} 1, \ \ \ & \text{if} \ \  f^{(12)}(j)=j,\\ 0, \ \ \ & \text{otherwise}\end{aligned}\right.\]
Determine $\sum_{f\in\mathbb X}\sum_{j=1}^nT_{f}(j).$
(Here $f^{(k)}(x)=f(f^{(k-1)}(x))$ for all $k\geq 2.$)
1. **Understanding the Problem:**
   We need to determine the sum \(\sum_{f\in\mathbb{X}}\sum_{j=1}^n T_f(j)\), where \(T_f(j)\) is defined as:
   \[
   T_f(j) = \begin{cases} 
   1 & \text{if } f^{(12)}(j) = j, \\
   0 & \text{otherwise}
   \end{cases}
   \]
   Here, \(f^{(k)}(x)\) denotes the \(k\)-th iterate of \(f\), i.e., applying \(f\) \(k\) times.

2. **Counting Functions with Cycles of Length 12:**
   We need to count the number of bijective functions \(f\) such that \(f^{(12)}(j) = j\) for a given \(j\). This means \(j\) must be part of a cycle of length 12 in the permutation \(f\).

3. **Cycle Lengths in Permutations:**
   In a permutation of \(n\) elements, the cycle containing \(j\) can have lengths that are divisors of 12 (i.e., 1, 2, 3, 4, 6, 12). We are interested in the case where the cycle length is exactly 12.

4. **Counting Permutations with a Cycle of Length 12:**
   To form a cycle of length 12 including \(j\), we need to choose 11 other elements from the remaining \(n-1\) elements. The number of ways to choose these 11 elements is \(\binom{n-1}{11}\). The number of ways to arrange these 12 elements in a cycle is \((12-1)!\) (since a cycle of length 12 has \((12-1)!\) distinct arrangements).

5. **Counting Remaining Elements:**
   The remaining \(n-12\) elements can be arranged in \((n-12)!\) ways.

6. **Total Number of Permutations with a Cycle of Length 12:**
   Therefore, the total number of permutations where \(j\) is in a cycle of length 12 is:
   \[
   \binom{n-1}{11} \cdot 11! \cdot (n-12)! = \frac{(n-1)!}{11!} \cdot 11! \cdot (n-12)! = (n-1)!
   \]

7. **Probability Calculation:**
   The probability that a randomly chosen permutation \(f\) has \(f^{(12)}(j) = j\) is:
   \[
   \mathbb{P}[T_f(j) = 1] = \frac{(n-1)!}{n!} = \frac{1}{n}
   \]

8. **Expected Value:**
   The expected value of \(T_f(j)\) for a fixed \(j\) is:
   \[
   \mathbb{E}[T_f(j)] = \mathbb{P}[T_f(j) = 1] = \frac{1}{n}
   \]

9. **Summing Over All \(j\):**
   Summing over all \(j\) from 1 to \(n\), we get:
   \[
   \sum_{j=1}^n \mathbb{E}[T_f(j)] = \sum_{j=1}^n \frac{1}{n} = 1
   \]

10. **Summing Over All \(f\):**
    Since there are \(n!\) permutations in \(\mathbb{X}\), the total sum is:
    \[
    \sum_{f\in\mathbb{X}}\sum_{j=1}^n T_f(j) = n! \cdot 1 = n!
    \]

The final answer is \(\boxed{n!}\)