An automorphism on generating set of a group Let $G$ be a finite group  and $A=\{a_{1},...a_{k}\}$ and $B=\{b_{1},...,b_{k}\}$ be two minimal generating sets of $G$  such that $|a_{i}| = |b_{i}|$ for $i=1,\dots,k$. We define $\alpha(a_{i})=b_{i}$. Is it an automorphism?
(At first I thank Derek Holt and  Tobias Kildetoft for their explanations.
Now I changed the conditon to minimal generating sets.
If $a_{1}^{t_{1}}...a_{k}^{t_{k}}=1 $ for $t_{i}<\mid a_{i}\mid$, then $t_{1}=t_{2}=...t_{k}=0$. Also we have this condition for the set B.)
Now I think with this condition added $\alpha$ is automorphism.
Thank you
 A: It's still not true even for abelian groups. Let $G = C_2 \times C_4 \times C_4 = \langle x \rangle \times \langle y \rangle \times \langle z \rangle$. Then $\{xy,y,z\}$ and $\{ xz,y,z\}$ are both minimal generating sets, but there is no automorphism $\alpha:xy \mapsto xz, y \mapsto y, z \mapsto z$, because it would have to map $x$ to $xy^{-1}z$, which has order $4$.
A: Here is a case where it does hold:
If $G$ is elementary abelian, then the generating sets correspond to bases over a suitable prime field (due to minimality), and the statement now just becomes the familiar argument from linear algebra that one can send any basis to any other basis via an invertible linear map.
A: Here is a generally true version: every group has a presentation where one lists generators and certain relations that they have amongst each other. Given any two generating subsets of the group that satisfy those defining relations, there is an automorphism that takes one set to the other.
For finite abelian groups one can always take the presentation to be a minimal generating set with the relations only the order relations and the obvious commutativity relations (which are automatic on any subset of the abelian group). Compare to Derek Holt's example: $C_2 \times C_4 \times C_4 = \langle x \rangle \times \langle y \rangle \times \langle z \rangle$ has presentation $$\left\langle x,y,z ~\middle|~ x^2 = y^4 = z^4 = 1, yx=xy, zx=xz, zy=yz \right\rangle$$
So absolutely any generating set $(a,b,c)$ with orders 2,4,4 will define an automorphism via $x\mapsto a, y\mapsto b, z \mapsto c$.
However, not all minimal generating sets satisfy these relations. For instance $a=xy,b=y,c=z$ and $a'=xz, b'=y, c'=z$ are both minimal generating sets with orders 4,4,4 but $a \mapsto a', b \mapsto b', c\mapsto c'$ runs into trouble as $x = ab^3 \mapsto a' (b')^3 = xy^3z$, but the first has order 2 and the second has order 4, an impossibility for a homomorphism. These generating sets don't satisfy the defining relations, so we get into trouble. The first set does satisfy these relations, but notice the extra bold one!
$$C_2 \times C_4 \times C_4 = \left\langle a,b,c ~\middle|~ a^4 = b^4 = c^4 = 1, \mathbf{a^2=b^2}, ba=ab, ca=ac,cb=bc \right\rangle$$
The generating set $(a',b',c')$ does not satisfy these relations (well not the bold one at least), so the automorphism does not exist.
This generator and relations method works for arbitrary groups as well. For instance the simple group of order 60 has a nice presentation $$\left\langle a,b ~\middle|~ a^2 = b^3 = (ab)^5 \right\rangle$$ where all the relations are orders, but not necessarily orders of generators. At any rate, any two distinct elements $a\neq b$ of any group that satisfy these relations generate a subgroup isomorphic to $A_5$, and given any two pairs $a\neq b$ and $a'\neq b'$ that both satisfy the relations, the function $a\mapsto a', b \mapsto b'$ defines an isomorphism between $\langle a,b\rangle$ and $\langle a',b'\rangle$.
