Let $a$ and $b$ $\in \mathbb C^N$ and $a[k] = a[k \mod N]$ (same with b). Then the circular convolution of $a$ and $b$ is defined by
$$ (a * b)[n] = \sum_{p=0}^{N-1} a[p] b[n-p].$$
I have a problem understanding the flip operation on the convolution. Let $(Ja)[n] = a[-n]$, then
$$ J(a*b)[n] = \sum_{p=0}^{N-1} a[p]b[-n-p].$$
so we first apply the convolution and then we flip the n - is this correct? Or do we flip the whole index of $b$, meaning that
$$ J(a*b)[n] = \sum_{p=0}^{N-1} a[p]b[-n+p].$$ On the other hand we have
$$((Ja)*(Jb))[n] = \sum_{p=0}^{N-1}a[-p]b[n+p].$$ so we first flip $a$ and $b$ and then apply the convolution. However, my book says $J(a*b)[n] = ((Ja)*(Jb))[n]$, which is not the case with my calculation, so I'm sure did something wrong. Can you help me finding the logical mistake here?