Does $$ \sum_{n=1}^\infty \frac{1}{\phi(n)^s}$$
have a euler product and functional equation? $\phi(n)$ is the euler phi function.
Since $\phi(n)$ is multiplicative I think the series could have a euler product and functional equation.
$$ \sum_{n=1}^\infty \frac{1}{\phi(n)^s}= \sum_{n=1}^\infty \frac{a(n)}{n^s}$$
where $a(n)$ is sequence https://oeis.org/A058277 in the OEIS.
I did some research and found many relations between the zeta function and other special functions such as:
$$ \sum_{n=1}^\infty \frac{\phi(n)}{n^s}=\frac{\zeta(s-1)}{\zeta(s)}. $$
Firstly, $$\phi(p^k)=p^k\left(1-\frac1p\right)\qquad{\text{$k\ge1$, prime $p$}}$$
Define $f(k)=\phi(k)^{-s}$.
For the moment, consider the set of integers $S_N=\{k\,|\,k=p_1^{a_1}p_2^{a_2}\cdots p_N^{a_N},a_{(\cdot)}\ge1\}$.
Then, $$\begin{align} \sum_{k\in S_N}f(k) &=\sum^\infty_{a_N=1}\cdots\sum^\infty_{a_1=1} \left[p_1^{a_1}\cdots p_N^{a_N} \left(1-\frac1{p_1}\right)\cdots\left(1-\frac1{p_1}\right)\right]^{-s} \\ &=\left(\prod^N_{i=1}\frac1{1-1/p_i}\right)^s\cdot\prod^N_{j=1}\sum^\infty_{a_j=1}p_j^{-a_js} \\ &=\prod^N_{i=1}\frac{(1-1/p_i)^{-s}}{p_i^s-1} \end{align} $$
Define $$f_i:=\frac{(1-1/p_i)^{-s}}{p_i^s-1}$$
Now we want to find $\displaystyle{\sum_{k\in S^*_N}f(k)}$ where $S^*_N=\{k\,|\,k=p_1^{a_1}\cdots p_N^{a_N},a_{(\cdot)}\color{red}{\ge0}\}$
Summing $f(k)$ over all the elements in $S_N^*$ with $a_\alpha=0$ and other indexes non-zero gives $\displaystyle{\frac1{f_\alpha}\prod^N_{i=1}f_i}$.
How about having two zero indexes $a_\alpha,a_\beta$? $\displaystyle{\frac1{f_\alpha f_\beta}\prod^N_{i=1}f_i}$. Three? $\displaystyle{\frac1{f_\alpha f_\beta f_\gamma}\prod^N_{i=1}f_i}$.
Summing all these and factorizing naturally give $$\sum_{k\in S^*_N}f(k)=\left(1+\frac1{f_1}\right)\left(1+\frac1{f_2}\right)\cdots\left(1+\frac1{f_N}\right)\cdot\prod^N_{i=1}f_i=\prod^N_{i=1}(1+f_i)$$
Taking the limit $N\to\infty$, we find that $$\sum^\infty_{n=1}\frac1{\phi(n)^s}=\prod_{\text{prime }p}\left(1+\frac{(1-1/p)^{-s}}{p^s-1}\right)$$
I am still working on the functional equation. It is clear that the function is holomorphic on $\text{Re }s>1$, and is likely to have a pole at $s=1$, as plugging in $s=1$ gives $\prod_p\left(1+\frac1p\right)=\infty$.
A few more words on analytic continuation:
Obviously, $$\begin{align} F(s):=\sum^\infty_{n=1}\frac1{\phi(n)^s} &=\prod_{\text{prime }p}\left(1+\frac{(1-1/p)^{-s}}{p^s-1}\right)\\ &=\prod_{p}\frac1{1-p^{-s}}\cdot\prod_p[1-p^{-s}+(p-1)^{-s}] \\ &=\zeta(s)\cdot \underbrace{\prod_p[1-p^{-s}+(p-1)^{-s}]}_{G(s)} \end{align} $$
Further analysis shows that $\text{Re }s=0$ is the natural boundary of $F(s)$.
Firstly, by means of successive series expansion, we obtain $$\begin{align} \ln(1-p^{-s}+(p-1)^{-s}) &=\sum^\infty_{n=1}\frac{\left[p^{-s}-(p-1)^{-s}\right]^n}{n} \\ &=\sum^\infty_{n=1}\frac{1}{n}\sum^n_{r=0}\binom nr p^{-s(n-r)}(-1)^r(p-1)^{-sr} \\ &=\sum^\infty_{n=1}\frac{p^{-ns}}{n}\sum^n_{r=0}\binom nr (-1)^r\left(1-\frac1p\right)^{-sr} \\ &=\sum^\infty_{n=1}\frac{p^{-ns}}{n}\sum^n_{r=0}\binom nr (-1)^r\sum^\infty_{k=0}\binom{-sr}{k}\frac{(-1)^k}{p^k} \\ &=\sum^\infty_{n=1}\sum^\infty_{k=0}\alpha_{n,k}(s)\frac1{p^{k+ns}} \end{align} $$
where $$\alpha_{n,k}(s)=\frac{(-1)^k}{n}\sum^n_{r=0}(-1)^r\binom nr \binom{-sr}{k}$$
Furthermore, notice that $$\alpha_{n,0}(s)=\frac1n\sum^n_{r=0}(-1)^r\binom nr=0$$
Therefore, $$\ln(1-p^{-s}+(p-1)^{-s})=\sum_{(n,k)\in\mathbb N^2}\alpha_{n,k}(s)\frac1{p^{k+ns}}$$ $$\begin{align} \implies \ln G(s) &=\sum_p \ln(1-p^{-s}+(p-1)^{-s}) \\ &=\sum_{(n,k)\in\mathbb N^2}\alpha_{n,k}(s)\zeta_{\mathbb P}(k+ns) \\ \end{align} $$ where $\zeta_{\mathbb P}$ is the prime zeta function.
It is well known that $\zeta_{\mathbb P}$ has the natural boundary $\text{Re }s=0$, because $\mathcal S$ (the set of singularities of $\zeta_{\mathbb P}$) clusters on the imaginary axis. Hence, obviously $G(s)$ cannot be analytically continued across $\text{Re }s=0$. A functional equation does not exist.
Meanwhile, we obtained a representation of $F(s)$ in terms of well-known functions: