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This Exercise is in analytic number theory textbook(Apostol)

Is there any arithmetical function $g$ which is a multiplicative function such that

$\sum_{k\leq n} f(gcd(n,k)) = \sum_{d|n} f(d)g(\frac{n}{d})$ for every arithmetical function $f$ ??(in deed, Textbook says me to prove existence, so it certainly exists.)

Let's think about it simply. That is,Consider that $n=p$ which $p$ is prime Then,

$\sum_{k\leq p} f(gcd(p,k)) = (p-1)f(1) + f(p)$ And $ \sum_{d|p} f(d)g(\frac{p}{d}) = f(1)g(p) + f(p)$

So, $g(p)=p-1$ where $f(1)$ is nonzero. Hence we can guess g(n) to be Euler totient function $\phi$. (Is my thinking reasonable?)

Next, Let $n=p^2$ Then, in the same way,

$\sum_{k\leq p^2} f(gcd(p^2,k)) = (p^2-2)f(1) +f(p)+ f(p^2)$ And $ \sum_{d|p^2} f(d)g(\frac{p^2}{d}) = f(1)g(p^2) +f(p)g(p)+ f(p^2)$

In this case, i can not think of $g $. as a Euler totient function. How should i do ?

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    Just saying: $$\sum_{k\leqslant n} f(\gcd(n,k)) = \sum_{d\mid n} \sum_{\gcd(n,k) = d} f(d) = \sum_{d\mid n} \sum_{\gcd\bigl(\frac{n}{d},\frac{k}{d}\bigr) = 1} f(d) = \sum_{d\mid n} \varphi(n/d)f(d)$$ – 2017-01-17
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    @DanielFischer Thanks for the explanation. Please explain the first equal sign. – 2017-01-17
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    We group the terms by $\gcd(n,k)$. Can you see why $$\sum_{k \leqslant n} f(\gcd(n,k)) = \sum_{g = 1}^n \Biggl(\sum_{\gcd(n,k) = g} f(g)\Biggr)\,?$$ Then one just drops the unnecessary terms for $g\nmid n$. – 2017-01-17
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    @DanielFischer yes, Thank you for your kind explanation. – 2017-01-17

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There is a mistake in your calculations for $p^2$. Note that not only $p$, $p^2$ are not coprime with $p^2$, but any $kp$, $k = 1, \ldots, p$ is. Hence $$ \sum_{k\le p^2} f(\gcd(k,p^2))= (p^2 -p)f(1) + (p-1)f(p) + f(p^2) $$ So the totient function is still a fine guess.