Let $~\rho(x)~=~\dfrac{\Gamma(x)\cdot\zeta(2x)}{\pi^x}.~$ Then $~\rho\Big(\tfrac14+x\Big)~=~\rho\Big(\tfrac14-x\Big)~$ constitutes the functional

  equation for the Riemann $\zeta$ function. The presence of the product $~\Gamma(x)\cdot\zeta(2x)~$ is perfectly

  understandable, inasmuch as the poles of the former coincide with the $($trivial$)$ zeroes of the latter;

  as is also the symmetry with regard to $\tfrac14,$ since this value stands midway between $\rho$'s two poles.

What poses serious difficulties from an intuitive perspective, however, is the presence of $~\dfrac{\zeta(2x)}{\pi^x}~$

instead of the expected $~\dfrac{\zeta(2x)}{\pi^{\color{red}2x}},~$ given the fact that $\zeta(2k)$ always possesses a known closed

  form in terms of $\pi^{\color{red}2k}$ rather than merely $\pi^k$ for integer values of the argument k. One can, of course,

  always write $~\rho(x/2)~=~\dfrac{\Gamma(x/2)~}{~\Gamma(1/2)^x}\cdot\zeta(x),~$ but, for all its niceness, the latter appears somewhat

  contrived, inasmuch as the power of $\pi$ is clearly a contribution of Riemann's $\zeta$ rather than Euler's

 $\Gamma$ function.

Let $\Lambda(s)=\pi^{-s/2} \Gamma\left(\frac{s}{2}\right)\zeta(s).$ Then $\Lambda\left(\frac{1}{2} + s\right) = \Lambda\left(\frac{1}{2} - s\right)$ constitutes the functional equation for the Riemann $\zeta$ function. The presence of the product $\Gamma\left(\frac{s}{2}\right) \zeta(s)$ is perfectly understandable, inasmuch as the poles of the former coincide with the (trivial) zeroes of the latter; as is also the symmetry with regard to $\frac{1}{2}$ since this value stands midway between the two poles of $\Lambda$.

What poses serious difficulties from an intuitive perspective, however, is the presence of $\pi^{-s/2} \zeta(s)$ instead of the expected $\pi^{-s} \zeta(s)$ given the fact that $\zeta(2k)$ always possesses a known closed form in terms of $\pi^{2k}$ rather than merely $\pi^k$ for integer values of the argument $k$. One can, of course, always write $\Lambda(s) = \Gamma\left(\frac{1}{2}\right)^{-s} \Gamma\left(\frac{s}{2}\right) \zeta(s)$ but, for all its niceness, the latter appears somewhat contrived, inasmuch as the power of $\pi$ is clearly a contribution of Riemann's $\zeta$ rather than Euler's  $\Gamma$ function.