Why real part of FFT converts image into rotation + original?
Let's say your image is given by $I(x,y)$. Then its Fourier transform is given by
$$ 
I^f(\omega_x,\omega_y) = \int_x\int_yI(x,y)e^{j\omega_xx}e^{j\omega_yy}dxdy
$$

Now you take the real part and perform the inverse:

\begin{align}
I_m(\alpha,\beta) &amp;= \int_{\omega_x}\int_{\omega_y}\Re\left\{I^f(\omega_x,\omega_y)\right\} e^{j\omega_x\alpha}e^{j\omega_y\beta}d\omega_xd\omega_y
\\
&amp;= \int_{\omega_x}\int_{\omega_y}\Re\left\{\int_x\int_yI(x,y)e^{j\omega_xx}e^{j\omega_yy}dxdy\right\} e^{j\omega_x\alpha}e^{j\omega_y\beta}d\omega_xd\omega_y\\
&amp;= \int_x\int_yI(x,y)\int_{\omega_x}\int_{\omega_y}\Re\left\{e^{j\omega_xx}e^{j\omega_yy}\right\} e^{j\omega_x\alpha}e^{j\omega_y\beta}d\omega_xd\omega_ydxdy
\end{align}

You can clearly see that the inner integral is the 2D Fourier transform of 
$$
\cos(\omega_xx)\cos(\omega_yy) + \sin(\omega_xx)\sin(\omega_yy)
$$
which is
$$
\frac{1}{2}\left[\delta(x-\alpha)\delta(y-\beta) + \delta(x+\alpha)\delta(y+\beta)\right]
$$

Substituting the result to $I_m$ yields
$$
I_m(x,y) = \frac{1}{2}\left[I(x,y)+I(-x,-y)\right] 
$$

Of course in your case $x,y&gt;0$, however the discrete Fourier transform assumes your signal is $N$-periodic and you get
$$
I_m(x,y) = \frac{1}{2}\left[I(x,y)+I(N-x,M-y)\right] 
$$
where $N,M$ are the dimensions of your image. I think you can see now why you got that result.