Let f, g: N → N such that f(n+1)=f(n)+f(1)

$f(n+1)-f(n)=f(1)$

$\Rightarrow \mathrm{f}(\mathrm{n})=\mathrm{nf}(1)$

$\Rightarrow \mathrm{f}$ is one-one

Now, Let $\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}_{2}\right)\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}_{1}\right)\right)$

$\Rightarrow \mathrm{g}\left(\mathrm{x}_{2}\right)=\mathrm{g}\left(\mathrm{x}_{1}\right)$ (as $\mathrm{f}$ is one-one)

$\Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}$ (as fog is one-one)

$\Rightarrow \mathrm{g}$ is one-one

Now, $\mathrm{f}(\mathrm{g}(\mathrm{n}))=\mathrm{g}(\mathrm{n}) \mathrm{f}(\mathrm{l})$

may be many-one if

$\mathrm{g}(\mathrm{n})$ is many-one