To find the number of bijections $f: A \to A$ that satisfy the given conditions, we can break the problem down into two distinct cases based on the logical biconditional statement $f(4) = 5 \iff f(5) = 4$.
The statement $P \iff Q$ is logically equivalent to saying that either both $P$ and $Q$ are true, or both $P$ and $Q$ are false. Therefore, we have two cases:
- Case $1: f(4) = 5$ and $f(5) = 4$
- Case $2: f(4) \neq 5$ and $f(5) \neq 4$
In both cases, the baseline restrictions $f(1) \neq 1$, $f(2) \neq 2$, and $f(3) \neq 3$ must still be satisfied.
Case $\mathbf{1}: f(4) = 5$ and $f(5) = 4$
Here, the mappings for $4$ and $5$ are fixed. The remaining elements in the domain are $\{1, 2, 3, 6\}$, and they must be mapped to the remaining available elements in the codomain, which are also $\{1, 2, 3, 6\}$.
We need to find the number of bijections on these $4$ elements such that $f(1) \neq 1$, $f(2) \neq 2$, and $f(3) \neq 3$. We can solve this using the Principle of Inclusion-Exclusion.
The total number of unrestricted permutations of the $4$ remaining elements is $4! = 24$.
The number of ways to violate exactly $1$ condition (e.g., $f(1)=1$) is $\binom{3}{1} \times 3! = 3 \times 6 = 18$.
The number of ways to violate exactly $2$ conditions is $\binom{3}{2} \times 2! = 3 \times 2 = 6$.
The number of ways to violate exactly $3$ conditions is $\binom{3}{3} \times 1! = 1 \times 1 = 1$.
Using Inclusion-Exclusion, the number of valid permutations for this case is:
$24 - 18 + 6 - 1 = 11$
Case $\mathbf{2}: f(4) \neq 5$ and $f(5) \neq 4$
In this case, we are looking for the number of bijections on the entire set $A$ of 6 elements, while simultaneously avoiding 5 specific mappings:
$f(1) \mapsto 1$
$f(2) \mapsto 2$
$f(3) \mapsto 3$
$f(4) \mapsto 5$
$f(5) \mapsto 4$
Notice that all $5$ of these forbidden mappings are independent. The inputs $\{1, 2, 3, 4, 5\}$ are all distinct, and the targeted restricted outputs $\{1, 2, 3, 5, 4\}$ are also all distinct. Because of this independence, we can apply the Principle of Inclusion-Exclusion directly to the $6$ elements with these $5$ restrictions.
The total number of unrestricted permutations of the $6$ elements is $6! = 720$.
The number of ways to violate $1$ condition is $\binom{5}{1} \times 5! = 5 \times 120 = 600$.
The number of ways to violate $2$ conditions is $\binom{5}{2} \times 4! = 10 \times 24 = 240$.
The number of ways to violate $3$ conditions is $\binom{5}{3} \times 3! = 10 \times 6 = 60$.
The number of ways to violate $4$ conditions is $\binom{5}{4} \times 2! = 5 \times 2 = 10$.
The number of ways to violate all $5$ conditions is $\binom{5}{5} \times 1! = 1 \times 1 = 1$.
Using Inclusion-Exclusion, the number of valid permutations for this case is:
$720 - 600 + 240 - 60 + 10 - 1 = 309$
Total Number of Bijections :
Since Case $1$ and Case $2$ are mutually exclusive and cover all possibilities of the biconditional statement, we simply add the valid permutations from both cases to get the final answer:
$$11 + 309 = \boxed{320}$$
There are $\mathbf{320}$ bijections that satisfy all the given conditions.
