The following series seems convergent for all $s\in \mathbb{C}$:

$$\displaystyle f(s):=\sum_{n=1}^\infty \frac{(-1)^n}{(n+s)^{n+s}}$$

The function itself does not appear to have any real or complex zeros, however numerical evidence suggests that all zeros of:

$$f(s) \pm f(1-s)$$

reside on the line $\Re(s)=\frac12$. Their density apparently increases in a quite regular manner.

1) Could this be proven? When using the finite series $\sum_{n=1}^N$, the claim seems to hold for each $N$.

2) Could there exist a functional equation between $f(s)$ and $f(1-s)$? The only relation I found so far is that $f(-1)=f(1)$.

**EDIT:** A partial result on the second question. Based on Joro's findings below and realising that for integers and half-integers an infinite number of terms could be cancelled out when adding or substracting $f(s)$ and $f(1-s)$, the following reflective formulae hold:

For all integers $s \in \mathbb{Z}$:

$$f(s)+f(1-s) +\sum_{n=0}^{2(s-1)} \frac{(-1)^{n}}{(s-n)^{s-n}} =0$$

For all half-integers $s+\dfrac{1}{2}$ with $s \in \mathbb{Z}$:

$$f(s)-f(1-s) +\sum_{n=0}^{2(s-1)} \frac{(-1)^{n}}{(s-n)^{s-n}} =0$$

The non-alternating series $\displaystyle g(s):=\sum_{n=1}^\infty \frac{1}{(n+s)^{n+s}}$ has one formula for integers and half-integers:

$$g(s)-g(1-s) +\sum_{n=0}^{2(s-1)} \frac{1}{(s-n)^{s-n}} =0$$