Maximum phase with the same magnitude response
Note that the zeros of the given transfer function have a negative real part:

$$z_{01}=\frac{1}{2\sqrt{2}}(-1 + j),\quad z_{02}=z_{01}^*\tag{1}$$

With $(1)$ you can write the transfer function as

$$H(z)=4(1-z_{01}z^{-1})(1-z_{01}^*z^{-1})\tag{2}$$

Reflecting a zero across the unit circle means replacing it by its conjugated inverse, i.e. $z_{01}\rightarrow 1/z_{01}^*$, from which you get

$$H_{max}(z)=k\left(z-\frac{1}{z_{01}}\right)\left(z-\frac{1}{z_{01}^*}\right)=\frac{k}{|z_{01}|^2}(1-z_{01}z)(1-z_{01}^*z)\tag{3}$$

The constant $k$ can be determined by evaluating $(2)$ and $(3)$ at a point on the unit circle $|z|=1$. If you choose $z=1$ you get

$$H(1)=4(1-z_{01})(1-z_{01}^*)\tag{4}$$

and

$$H_{max}(1)=\frac{k}{|z_{01}|^2}(1-z_{01})(1-z_{01}^*)\tag{5}$$

Comparing $(4)$ and $(5)$ you get

$$\frac{k}{|z_{01}|^2}=4\tag{6}$$

and with $(3)$

$$H_{max}(z)=4(1-z_{01}z)(1-z_{01}^*z)\tag{7}$$

Comparing $(7)$ with $(2)$ you can see that

$$H_{max}(z)=H(z^{-1})=z^2+2\sqrt{2}z+4\tag{8}$$

In the time domain this corresponds to a time reversal. If $h[n]$ is the impulse response corresponding to $H(z)$, and $h_{max}[n]$ corresponds to $H_{max}(z)$, then the following holds:

$$h_{max}[n]=h[-n]\tag{9}$$