$\def\F\{\operatorname F}\def\E{\operatorname E}$
Elliptic F$(x,m)$ appears in elliptic E$(x,m) =\int_0^x\sqrt{1-m\sin^2(x)}dx$ parameter $m$ transformations. Luckily the DLMF has reciprocal/imaginary modulus transformations and Gauss/Landen transformations, but these both use $\E(x,k)= \int_0^x\sqrt{1-k^2\sin^2(x)} dx$. The following formulas work around $z=0$:
Reciprocal parameter transformation:
$$\E(z,m)=\sqrt m\E\left(\sin^{-1}\left(\sqrt m\sin(z)\right),\frac1m\right)+\frac{1-m}{\sqrt m}\F\left(\sin^{-1}(\sqrt m\sin(z),\frac1m\right)\tag1$$
Complementary parameter/imaginary argument transformation:
$$\E(z,m)=i\E(i\tanh^{-1}(\sin(z)),1-m)-i\F(i\tanh^{-1}(\sin(z)),1-m)+\sin(z)\sqrt{\frac{1-m\sin^2(z)}{\cos^2(z)}}$$
Negative parameter transform from ResearchGate using Elliptic $\E(z)$ and complex conjugate $\bar z$
$$\E(z,m)\mathop=^{m<1}\sqrt{1-m}\left(\E\left(\frac m{1-m}\right)-\E\left(\frac\pi 2-z,\frac m{1-m}\right)\right)\mathop=^{m>1}-\overline{\sqrt{1-m}\left(\E\left(\frac m{1-m}\right)-\E\left(\frac\pi 2-z,\frac m{1-m}\right)\right)}\tag2$$
Ascending Landen transformation:
$$\E(z,m)=(1+\sqrt m)\E\left(\frac12(\sin^{-1}(\sqrt m\sin(z))+z),\frac{2\sqrt m}{m+1}\right)+ (1-\sqrt m)\F\left(\frac12(\sin^{-1}(\sqrt m\sin(z))+z),\frac{2\sqrt m}{m+1}\right)-\sqrt m\sin(z)$$
Descending Landen transformation:
$$\E(z,m)=\frac{\sqrt{1-m}+1}2\E\left(\tan^{-1}(\sqrt{1-m}\tan(z))+z,\left(\frac{\sqrt{1-m}-1}{\sqrt{1-m}+1}\right)^2\right)-\sqrt{1-m}\F(z,m)+\frac{m\cos(z)\sin(z)}{2\sqrt{1-m\sin^2(z)}}$$
These formulas answer most of the original question, but they use $\frac{m\cos(z)\sin(z)}{2\sqrt{1-m\sin^2(z)}}$ and other complicated trigonometric expressions.
What are other simpler transformations for $\E(z,m)$, perhaps without trigonometric expressions like in $(1),(2)$?
(This is not an answer but an extended comment)
Note for example, the following transformation (using Mathematica notation). Suppose that $0<m<1$, then
$$E(\varphi,-m) = \sqrt{1+m}E\left(\varphi,\frac{ m}{m+1}\right)- \frac{m\sin(\varphi)\cos(\varphi)}{\sqrt{1+m\cos^2\varphi}}$$
For convenience put $\displaystyle \varphi = \frac{\pi}{2}$ and $\displaystyle m=\frac{1}{2}$
$$E\left(\frac{\pi}{2},-\frac{1}{2}\right) =\sqrt{\frac{3}{2}}E\left(\frac{\pi}{2},\frac{1}{3}\right)$$
So, in Mathematica notation we are transforming a negative modulus to a positive modulus while using the standard notation we are transforming a purely imaginary modulus to a positive one:
$$E\left(\frac{\pi}{2},\frac{i}{\sqrt{2}}\right) = \sqrt{\frac{3}{2}}E\left(\frac{\pi}{2},\frac{1}{\sqrt{3}}\right) $$
My advice is to completely ignore the Wolfram's notation for the theory of elliptic integrals and use it only for computation. They also put a warning on this issue in MathWorld.