Fixed point of Riemann Zeta function

I do not think your statement about fixed points in the plane to be true - it may be true for $Re(z)>1$ in the sense of being just one fixed point there, but otherwise $(s-1)\zeta(s)$ is an entire function of order 1 and maximal type (by the usual properties of the critical strip zeros - eg their ~$T\log(T)$ density and general stuff about entire functions of finite order - the usual notion of density of zeros for entire functions and the one for $\zeta$ differ a little but they have the same order of magnitude) and subtracting a polynomial like $s(s-1)$ doesn't change order 1 or maximal type as those depend on the Taylor coefficients at infinity for any entire function, so in particular $(s-1)\zeta(s) - s(s-1)$ is entire of order 1 and maximal type and those have lots of zeros - either they have density growing faster than T at infinity or the conditional sum of their reciprocals is not convergent by a theorem of Lindelof. Maximal type is crucial because obviously exponentials of linear polynomials have order 1 and arbitrary finite type.

Note that the reciprocal of the Gamma function is order 1 and maximal type but has the density of zeros ~T (say on the disk of radius T centered at the origin) as its zeros are just the negative numbers (so, in particular, the conditional sum of their reciprocals is not convergent, so it's possible the number of fixed points of $\zeta$ to be of order T only sure; similar considerations apply to any equation of the type $\zeta(s)=Polynomial(s)$ by multiplying with s-1 and reducing to considerations about entire functions of order 1 and maximal type.

Hmm...i did a run in my computer cause i found you question of fixed points interesting so..

the only result i got is this for $a=1.8337719154395\cdots$ and for $b=0$

$\zeta(1.8337719154395\cdots)=1.8337719154395\cdots$

wish you all the luck

note: this is an amateurs approach i'm no mathematician