Let $m$ be the smallest such element element in $\mathbb{N} $ such that $f(x)=x^m$ for all $x\in G$ . Such a element always exists as $\mathbb{N}$ is well ordered . Let $|G|=d$ . We will prove that $d | m $ i.e $d$ divides $m$ . Assume otherwise . So $m=dk+r $ where $0<r \leq d-1$( by division algorithm ) . Then $f(x)=x^m=x^{( dk+r)}=\{x^d\}^k x^r=x^r \forall x\in G$ .Contradicting minimality condition of $m$ . So $d$ divides $m$ . Hence $m=dk$ . Now take $x,y\in G$ . $f(xyx^{-1}y^{-1})=(xyx^{-1}y^{-1})^m= \{{(xyx^{-1}y^{-1})^d}\}^k=1$ . So $ xyx^{-1}y^{-1} \in ker f=\{1\} $ as $f$ is automorphism . So $xyx^{-1}y^{-1}=1 $ i.e $xy=yx $ . As $x, y$ was chosen arbitrary element , this is true for all elements of $G $ . Hence $G$ is commutative .

Your solution is very nice and elegant, but unfortunately it is also wrong. If $m<d$ (and this is always the case when we choose to minimize $m$), then $r=d$ and it does not contradict the minimality. It can actually be easily shown that $(m,d)=1$

It's a really nice problem. The trick is to consider a weaker condition: demand only that the inner automorphisms of $G$ are of the desired form. This turns out to be sufficient and has a big advantage that this property is inherited by subgroups and quotients, which allows a proof by induction on $|G|$. Another way is to use the structure theorem for Hamiltonian groups, but that's quite an overkill (however it allows to handle infinite groups as well). Btw with the problem as stated even more is true: $G$ must be cyclic (however with the condition only for inner automorphisms all we can say is that $G$ is abelian).

Hi, @GreenKeeper. Thank you for your insightful comments. If you have time, could you please write a complete solution? (for the sake of completeness)

Yes you are right . My solution is (half) wrong . I'm working on it . I think i made an interesting observation . I think $m$ must be a prime number . My reasoning is that if it were composite then for a prime $p$ we have $p|m$ .Then by Sylow's theorem there exists a non-identity element $a\in G $ such that $ |a|=p$ . Then $f(a)=a^m=\{a^p\}^{\frac{m}{p}}=1=f(1)$ . As $f$ is automorphism , $a=1 $ contradicting that $a$ was chosen as non-identity . What are your thoughts about it ?

Whether u and v, two elements in G,such that ord(u)| ord(v).Whether f:G->G, f(x)=v*x*v^-1, f(x)=x^m, x in G. Obtain f(v)=v=v^m, so v^(m-1)=e, so ord(v) |m-1. Result u ^(m-1)=e,so u=u^m=f(u)=v*u*v^-1.Finally,v*u=u*v.I'm not sure that is complete solution.

@alphaZ, this is also wrong. $p$ divides $m$, but not necessarily $d$ (the order of the group) and we can not use Sylow/Cauchy theorem.

Yes you are right . Looks like sleep deprivation is catching up to me . Anyway , can you show why $(m, d)=1 $

My solution is correct? Probably incomplete?

Assume $(m,d)\geq2$. Then there exists a prime $q$ so that $q\mid(m,d)$. We have $m=qt$, where $t\in\mathbb{N}^{\star}$. Take $x \in G$ an element of order $q$. But then $f(x)=x^{m}= (x^{q})^{t}=e^{t}=e$. Because $f$ is injective, we get $x=e$, which is false. @soryn, obviously, your solution is incomplete. But the idea is useful. I will post the official solution, which uses it.

Thanks, Mefistocle!

We know that $G/Z(G)$ and $Inn(G)$ are isomorphic , also there is a theorem that say "When $G/Z(G)$ is cyclic , G is abelian" so we can prove $Inn(G)$ is cyclic and its easy.

Here is the official solution.

I am curious to see other solutions too. This problem belongs to Marian Andronache.

OK, here's my solution. For any $g\in G$ let $f_g$ be the inner automorphism of $G$ induced be $g$, that is $f_g(x)=gxg^{-1}$ for $x\in G$. Let $(*)$ be a property that every inner automorphism $f_g$ of $G$ is of the form $f_g(x)=x^{m}$ for some $m\in\mathbb{Z}$ (for finite groups it doesn't make a difference whether or not we allow $m$ to be non-positive but it's perhaps more natural to do so). We'll prove by induction on $|G|$ that every finite group $G$ satisfying $(*)$ is abelian. For $|G|=1$ it's obvious. Now assume that $|G|>1$ and that the claim holds for all groups with smaller cardinality, and furthermore assume for contradiction that $G$ is not abelian. It's clear that $G$ is a Dedekind group, that is all subgroups of $G$ are normal. Also it's easy to see that all subgroups and all quotients of $G$ satisfy $(*)$ (this is the reason we only consider inner automorphisms, because we don't apriori know this for the non-inner ones), so by the induction hypothesis it follows that all proper subgroups of $G$ and all quotients of $G$ by non-trivial subgroups are abelian. Let $p$ be any prime dividing $|G|$ and let $H\le G$ be any subgroup of $G$ of order $p$, at least one such exist for every $p$ by Sylow. Since $G$ is non-abelian, $[G,G]$ is non-trivial. But $G/H$ is abelian, so $[G,G]\subseteq H$, and thus $[G,G]=H$. But this holds for every $p$ and every $H$, thus it follows that $G$ is a $p$-group for some prime $p$ and it has an unique subgroup of order $p$, which is equal to $[G,G]$. Since $Z(G)$ is non-trivial, we also have $[G,G]\subseteq Z(G)$. Since $G$ is non-abelian, there exist $g,h\in G$ that don't commute. Since $\langle g,h\rangle$ is a non-abelian subgroup of $G$, it follows $G=\langle g,h\rangle$. Let $[g,h]=ghg^{-1}h^{-1}$ be the commutator of $g$ and $h$. Since $[g,h]\in Z(G)$, we have $f_g([g,h])=[g,h]$. Also $f_g(h)=ghg^{-1}=[g,h]h$. It follows by induction that for every $k\in\mathbb{N}$ it's $f_g^k(h)=[g,h]^kh$. In particular, $f_g^p(h)=[g,h]^ph=h$. Also $f_g(g)=g$, thus $f_g^p(g)=g$ and since $G=\langle g,h\rangle$, it's $f_g^p=\mathrm{id}_G$ and thus the order of $f_g$ in $\mathrm{Inn}(G)$ is $p$ (it's not $1$ since $f_g(h)\ne h$). By symmetry, the order of $f_h$ is also $p$. Using once again $G=\langle g,h\rangle$, we have $\mathrm{Inn}(G)=\langle f_g,f_h\rangle$. Since $\mathrm{Inn}(G)\cong G/Z(G)$ is a quotient of $G$ by a non-trivial subgroup, $\mathrm{Inn}(G)$ is abelian. Altogether we get that $\mathrm{Inn}(G)$ is an abelian group generated by two elements of order $p$, so it's isomorphic to either $\mathbb{Z}/p$ or $\mathbb{Z}/p\times\mathbb{Z}/p$. But $G/Z(G)$ can never be a non-trivial cyclic group (for any group $G$), and thus $\mathrm{Inn}(G)\cong\mathbb{Z}/p\times\mathbb{Z}/p$. Now let $e$ be (the smallest) exponent of $G$. Then it's clear that for every inner automorphim $f_g$ the corresponding $m$ is determined uniquely modulo $e$, and that $(e,m)=1$ (otherwise the map $x\to x^m$ wouldn't be injective). So we have a well-defined map $\mathrm{Inn}(G)\to(\mathbb{Z}/e)^\times$, and it's straightforward to see that it's an injective homomorphism, and so $\mathrm{Inn}(G)$ embeds into $(\mathbb{Z}/e)^\times$ (which also gives an alternative proof that $\mathrm{Inn}(G)$ is abelian). Since $G$ is a $p$-group, we have $e=p^t$ for some $t\in\mathbb{N}$. If $p\ne2$ or $p=2$ and $t\le2$, then $(\mathbb{Z}/e)^\times=(\mathbb{Z}/p^t)^\times$ is a cyclic group of order $\varphi(p^t)=p^{t-1}(p-1)$, which gives a contradiction because a non-cyclic group $\mathbb{Z}/p\times\mathbb{Z}/p$ cannot embed into a cyclic one. Thus $p=2$ and $t\ge3$, then $(\mathbb{Z}/e)^\times=(\mathbb{Z}/2^t)^\times\cong\mathbb{Z}/2\times\mathbb{Z}/2^{t-2}$, which contains a unique subgroup isomorphic to $\mathbb{Z}/2\times\mathbb{Z}/2$ (formed be the four elements of order $1$ or $2$). We can specifically write down this subgroup as $\{[1]_{2^t},[-1]_{2^t},\big[5^{2^{t-3}}\big]_{2^t},\big[-5^{2^{t-3}}\big]_{2^t}\}$, but the only thing we need to know is that $[-1]_{2^t}$ belongs to this group. This is because it tells us that the map $x\to x^{-1}$ is an (inner, but we don't need that) automorphism of $G$, which means that $G$ is abelian, and that's once again a contradiction. This completes the proof. Obviously all the finite abelian groups satisty $(*)$. However, if we assume that all the automorphisms of $G$ are of the given form, then $G$ must be in fact cyclic. Indeed, the condition implies that all the subgroups of $G$ are characteristic. We already know that $G$ is abelian, thus $G$ is a product of cyclic groups of prime-power order. If $G$ wasn't cyclic, then it'd contain two summands $\mathbb{Z}/p^a$ and $\mathbb{Z}/p^b$ for some prime $p$ and $a\le b$. Then the map $([x]_{p^a},[y]_{p^b})\to([x+y]_{p^a},[y]_{p^b})$ is an automorphism of $\mathbb{Z}/p^a\times\mathbb{Z}/p^b$, which can be extended to an automorphism of $G$ by taking identity on the other summands. This show that the summand $\mathbb{Z}/p^b$ is a non-characteristic subgroup of $G$, a contradiction (or alternatively we can directly see that this automorphism of $G$ is not of the given form). Hence $G$ is cyclic.

If we don't assume that $G$ is finite, then it's still true that if $G$ safisties $(*)$ then it's abelian. This is because we still know that it's a Dedekind group, and if it was Hamiltonian (that is non-abelian Dedekind) then it's known that $G\cong Q_8\times A\times B$, where $Q_8$ is the quaternion group, $A$ is a sum of copies of $\mathbb{Z}/2$ (number of summands can have any cardinality, finite or infinite) and $B$ is an abelian group such that all its elements have a finite odd order. Since $Q_8$ has exponent $e=4$, and $(\mathbb{Z}/4)^\times=\{[1]_4,[-1]_4\}$, the only possible automorphisms of $Q_8$ of the form $x\to x^m$ are the identity and $x\to x^{-1}$. But the latter isn't a homomorphism since $Q_8$ isn't abelian, so that leaves only the identity. Now $Z(Q_8)=\{\pm1\}$, so $Q_8$ has three inner automorphisms that aren't of this form, given by the conjugation by $\pm i$, $\pm j$ and $\pm k$. Any of these can be extended to an inner automorphisms of $G$ by taking identity on $A$ and $B$, which gives a contradiction. Thus any (finite or infinite) group satisfies $(*)$ iff it's abelian. This raises the question which infinite groups have all the automorphisms of the given form. Here it matters whether we allow $m\in\mathbb{Z}$ or demand $m\in\mathbb{N}$. The infinite cyclic group $\mathbb{Z}$ has two automorphisms, the identity and $x\to-x$. So it satisfies the condition for $m\in\mathbb{Z}$ but not for $m\in\mathbb{N}$. I don't know whether there exist any non-cyclic infinite group that satisfies this either for $m\in\mathbb{Z}$ or even for $m\in\mathbb{N}$. EDIT: I've though about it a little more and it turns out the the answer is yes for $m\in\mathbb{N}$ and no for $m\in\mathbb{Z}$. The map $x\to x^{-1}$ is an automorphism of any abelian group, so if we demand that it's of the form $x\to x^m$ for $m\in\mathbb{N}$ then $G$ has a finite exponent $m+1$. But an abelian group with finite exponent is a sum of finite cyclic groups of bounded order by the first Prüfer theorem, so if $G$ is infinite then some summand $\mathbb{Z}/n$ must appear infinitely many times (we may assume that $n$ is a prime powers but it's not necessary) which gives rise to infinitely many automorphisms that are not of the required form. However there exist a lot of infinite abelian groups such that their only automorphisms are the identity and the inverse, see for example here, all of them obviously satisfy the condition with $m\in\mathbb{Z}$.

Really romanian maths olympiad has group theory. Wow and it is for high school students what is the curriculum of school there?