Why this two FIR filters have the same amplitude but different phase response
If you have a causal length $N$ FIR filter with impulse response $h[n]$, and with frequency response $H(e^{j\omega})$, and you invert it on the time axis, you get a new filter

$$g[n]=h[-n]\tag{1}$$

This filter is non-causal, but since it is an FIR filter, it can be made causal by shifting it to the left by $N-1$ samples:

$$f[n]=g[n-(N-1)]=h[N-1-n]\tag{2}$$

This is what you're doing. Now since

$$H(e^{j\omega})=\sum_{n=-\infty}^{\infty}h[n]e^{-jn\omega}\tag{3}$$

you have

$$G(e^{j\omega})=\sum_{n=-\infty}^{\infty}h[-n]e^{-jn\omega}=\sum_{n=-\infty}^{\infty}h[n]e^{jn\omega}=H(e^{-j\omega})$$

and

$$F(e^{j\omega})=G(e^{j\omega})e^{-j(N-1)\omega}=H(e^{-j\omega})e^{-j(N-1)\omega}$$

From (3) you can see that for real-valued $h[n]$ you have $H(e^{-j\omega})=H^*(e^{j\omega})$ (where $*$ denotes complex conjugation). So you finally get

$$F(e^{j\omega})=H^*(e^{j\omega})e^{-j(N-1)\omega}\tag{4}$$

and since $|e^{-j(N-1)\omega}|=1$, and $|H^*(e^{j\omega})|=|H(e^{j\omega})|$ you have

$$|F(e^{j\omega})|=|H(e^{j\omega})|\tag{5}$$

So the magnitude responses of the original filter and the time-reversed filter are identical. The phase responses are different: from (4) you have

$$|F(e^{j\omega})|e^{j\phi_F(\omega)}=|H(e^{j\omega})|e^{-j\phi_H(\omega)}e^{-j(N-1)\omega}\tag{6}$$

from which

$$\phi_F(\omega)=-\phi_H(\omega)-(N-1)\omega\tag{7}$$

follows. Since the group delay is the negative derivative of the phase, the group delays are related by

$$\tau_F(\omega)=-\tau_H(\omega)+N-1\tag{8}$$