Let $A=\left\{1,2,3,\dots ,10\right\}$ and $f:A\to A$ be defined as

$f\left(k\right)=\left\{\begin{array}{cl}k+1& \mathrm{if} k \mathrm{is}odd\\ k& \mathrm{if} k \mathrm{is}even\end{array}\right.$

Then the number of possible functions $g:A\to A$ such that $gof=f$ is:

$f\left(x\right)=\left\{\begin{array}{cl}x+1& \mathrm{if} x \mathrm{is}odd\\ x& \mathrm{if} x \mathrm{is}even\end{array}\right.$

$\because g:A\to$ A such that $g\left(f\left(x\right)\right)=f\left(x\right)$

$\Rightarrow$ If $x$ is even then $g\left(x\right)=x \dots \left(1\right)$

If $x$ is odd then $g\left(x+1\right)=x+1 \dots \left(2\right)$

from $\left(1\right)$ and $\left(2\right)$ we can say that $g\left(x\right)=x$ if $x$ is even

$\Rightarrow$ If $x$ is odd then $g\left(x\right)$ can take any value in set $A$

so number of $g\left(x\right)={10}^5\times 1$