Let $$A = \{1, 2, 3, \ldots, 10\}$$ and $$f : A \to A$$ be defined as
$$f(k) = \begin{cases} k + 1 & \text{if } k \text{ is odd} \\ k & \text{if } k \text{ is even} \end{cases}$$
Then the number of possible functions $$g : A \to A$$ such that $$gof = f$$ is:

The function $$f : A \to A$$ is defined as $$f(k) = k + 1$$ if $$k$$ is odd, and $$f(k) = k$$ if $$k$$ is even. So: $$f(1) = 2$$, $$f(2) = 2$$, $$f(3) = 4$$, $$f(4) = 4$$, $$f(5) = 6$$, $$f(6) = 6$$, $$f(7) = 8$$, $$f(8) = 8$$, $$f(9) = 10$$, $$f(10) = 10$$.

The range of $$f$$ is $$\{2, 4, 6, 8, 10\}$$. We need $$g \circ f = f$$, i.e., $$g(f(k)) = f(k)$$ for all $$k \in A$$.

This means $$g(2) = 2$$, $$g(4) = 4$$, $$g(6) = 6$$, $$g(8) = 8$$, and $$g(10) = 10$$. These five values of $$g$$ are completely determined.

However, the values $$g(1)$$, $$g(3)$$, $$g(5)$$, $$g(7)$$, and $$g(9)$$ are not constrained by the condition $$g \circ f = f$$ (since 1, 3, 5, 7, 9 are not in the range of $$f$$). Each of these can be any element of $$A = \{1, 2, \ldots, 10\}$$.

Therefore the number of possible functions $$g$$ is $$10 \times 10 \times 10 \times 10 \times 10 = 10^5$$.