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A necessary condition for an example is that there are positive integers $a,b$ such that $\prod_{i=0}^{2N-1}(a+ib)$ is a square. Differently phrased (divide by $b^{2N}$), we ask for rational solutions …

Here is a more pedestrian version of Fedor Petrov's argument: We set $\lambda=1/\log6$ and claim that $(-1)^n(1/\zeta)^{(n)}(\lambda n)>0$ if $n$ is big enough. With $a_k=\log k/k^\lambda$, we have \b …

The answer is $\varepsilon(n)=1-\frac{2}{n}$. Clearly, $\lvert\varphi_p(A)\rvert\ge2$ for all primes $p$. However, for every $n\ge2$ there is a set $A$ of size $n$ such that $\lvert\varphi_p(A)\rvert= …

Maybe I misunderstand the question, but doesn't the set $A=\{i\cdot n!\,|\,1\le i\le n\}$ have $\lvert\varphi_p(A)\rvert=1$ for all $p$ for which $\varphi_p$ is not injective on $A$?

I think both congruences are essentially trivial: As $\varphi(4n+3)$ is even and $(4n+3)-1$ is $2$ times an odd number, the divisors which are counted come in pairs $t,2t$, where $t$ is odd. Similarl …

The answer is yes, for further infos see the references given at the On-Line Encyclopedia of Integer Sequences.