Automorphism acting trivially on more than half the elements of the group is trivial Let $G$ be a group of finite order, wih  $\phi:G\rightarrow G$ an automorphism on $G$. 
Assume $A=$ {$ g\in G|\phi(g)=g $} consists of more than half of the elements of $G$. 
In other words, $|A| > |G|/2$.
Prove $\phi$ is the identity automorphism.
 A: As Wojowu commented, one way to show this is by observing that $A$ is a subgroup of $G$ and then applying Lagrange's theorem. However, we don't need to appeal to Lagrange (or a bit more fairly, basically the same idea behind the proof of Lagrange's theorem works perfectly well here). Specifically, we can argue as follows:

Suppose $g\not\in A$. Consider the set $gA:=\{gh: h\in A\}$. Now since the group operation is left-cancellative, the map $h\mapsto gh$ is a bijection between $A$ and $gA$. Meanwhile, for $gh=l\in gA$, we have $\phi(l)=\phi(g)\phi(h)=\phi(g)h\not=gh$ since $\phi(g)\not=g$ and the group operation is right-cancellative. Putting this together we get $$\vert G\vert\ge\vert A\cup gA\vert = \vert gA\vert+\vert A\vert> {2\vert G\vert\over 2}=\vert G\vert,$$ a clear contradiction.

Note that this stripped-down argument also has a neat bit of extra generality. Suppose I have a finite magma $M$ and an automorphism $\phi$ fixing more than half of $M$. We do not need to have $\phi$ be the identity, and this is a good exercise:

 Consider, for any $n$, the set $X=\{c_1,...,c_n\}\cup\{a,b\}$ equipped with the binary operation $*$ acting as left projection on $\{c_1,...,c_n\}$ and on $\{a,b\}$ and otherwise satisfying $a*c_i=a=c_i*a$ and $b*c_i=b=c_i*b$. Then the map swapping $a$ and $b$ and fixing each $c_i$ is an automorphism of $M$. Now pick a truly gargantuan $n$, like $5$.

But we do get something, namely that the set $A$ of fixed points of a $\phi$ which fixes more than half of $M$ has a certain "closure" property. Suppose $g\in M$ has the following property (a kind of "local cancellation over $A$"): for all $h_1,h_2\in A$, we have $gh_1=gh_2\rightarrow h_1=h_2$ and $gh_1=h_2h_1\rightarrow g=h_2.$ Then $g\in A$.
(And of course $A$ must also be a sub-magma, but that's just a general structural fact.)
Of course, this is (so far as I know) utterly useless - but it's still a good exercise in extracting some generality from a "barer" argument.
