If $A_1$ and $A_2$ are unitary matrices ,and if $CA_1C^{-1}=A_2$,then is $C$ is also unitary? $A_1$ and $A_2$ are matrix of the same unitary transformation, w.r.t different orthonormal basis. And $C$ is the transition matrix(hence, $CA_1C^{-1}=A_2$).
To show if $A_1$ and $A_2$ are unitary matrices then $C$ is also unitary.
where unitary means $AA^*=A^*A=I$, $A^*$ is the adjoint of $A$.
I have been trying this , but could not come with proof.
 A: The edits really improved the question. Now the answer is pretty much already in the way you state it.
EDIT: the last comment by user1551 made me realize that the matter is not quite as simple as I thought. Something in below answer is wrong but I am not quite sure what or where. PROCEED WITH CAUTION! (End of edit)
In the new version $A_1$ describes linear transformation $A$ w.r.t. orthonormal basis $B_1$ and $A_2$ describes the same transformation $A$ w.r.t. basis $B_2$. This means that we know what $C$ does: it maps basis $B_1$ to basis $B_2$!
Now unitary means 'preserves the inner product' which in human language means 'preserves angles and distances'. It is pretty obvious that $C$ does indeed do this for the elements of $B_1$: they all have length 1 and their images are elements of $B_2$ so these images also have length one. Similar the angle between two elements of $B_1$ equals the angle between their images under $C$, namely 90 degrees (or $\pi/2$ radians).
The question then remains: 'okay, $C$ preserves angles and distances (or more general: the inner product) when applied to the finitely many elements of the basis $B_1$ but does that imply that it does the same when applied to general vectors?'
And the answer is 'well yeah, that is (more or less) what $B_1$ being a basis means'.
(This way of formulating things may feel a bit too informal for your taste, but it can easily be formalized when we accept the definition of unitary as 'preserving the inner product'. The link with the definition $A^*A = I$ is then that the inner product of $u$ and $v$ in this language is $u^*v$ so that the inner product of $Au$ and $Av$ becomes $(Au)*(Av) = (u^*A^*)(Av) = u^*(A^*A)v^*$, which equals $uv$ for all $u$ and $v$ if and only if $A^*A = I$.)
