# Automorphism of $D_8$ [duplicate]

I am trying to prove that $$Aut(D_8) \equiv D_8$$. It is not hard to see that $$\lvert Aut(D_8)\rvert = 8$$. Indeed, it is at most $$8$$ as $$r$$ (canonical rotation) has order $$4$$ and $$s$$ (canonical reflection) has order $$2$$. On the other hand, $$D_8$$ is normal in $$D_{16}$$ which acts by conjugation on $$D_8$$. So, the order is exactly $$8$$. However, I am having troubles to show that $$Aut(D_8) \equiv D_8$$ (I could check all possible solutions and stop once I have 8 different automorphisms but I wonder if there is a simpler solution).
## marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 12 '18 at 9:30
• You can do it by showing that there are $f,g\in \text{Aut}(D_8)$ such that $f^2=1,g^4=1$ and $fgf^{-1}=g^{-1}$. – user9077 Dec 12 '18 at 0:55
You already know it's an order 8 factor group given by $$D_{16}$$ mod its center. Clearly that has 2 elements of order 4 (images of the order four order 8 elements) and 5 of order 2. The only such group is $$D_8$$....just list all 5 groups of order 8 and how many elements of each order they each have.