I had written up a hint in the comments, I'll make that into an answer:
Try to write $f$ in cycle notation. What happens to the identity under $f$? And, pick a $g\in G$ and since $a$ is not the identity, you'll have cycles of length $2$ and there will be $\%@\%@$ of them. So...
My definition of odd Permutation
A permutation of $G$ is said to be odd if there are odd number of transpositions in some cycle notation of $f$.
Note: This requires some proof that if one cycle notation of $f$ has odd number of transpositions, so will any other cycle. (This can be thought of as a consequence of the fact that identity is a even transposition, which in turn will require a proof.)
Here's an expanded version:
Note that since $a$ is an element of order $2$, it cannot be the identity. Hence, the permutation is not identity on $G$. Now note that no element is fixed. Why? $ag=g \iff a=e_G$. This is another way of saying, that left multiplication is a transitive action. This is another way of saying that left multiplication is a Quasi Regular action. (Thanks to Schmidt for bringing to my notice an abuse of terminology through the comments!)
So, all elements end up in disjoint cycles of order $2$. So, the $f$ in cycle notation is a product of exactly $2n+1$ cycles!
Hence $f$ is an odd permutation. $\blacksquare$
If it helps, we can take an example group:
- In $\mathbb Z_6$. Let $a = \bar 3$. Note that, since this is an Abelian group, $f(g)= \overline{ \bar a +\bar g}$.
By working things out, we'll see that $f \equiv (\bar 0, \bar 3)(\bar 1, \bar 4)(\bar 2, \bar 5)$
- Similarly for $\mathbb Z_{10}$, and $a= \bar 5$
Note that, $f \equiv (\bar 0, \bar 5 )(\bar 1, \bar 6)(\bar 2, \bar 7)(\bar 3, \bar 8)(\bar 4, \bar 9 )$