Is it possible to recognize when an endomorphism of a finite dimensional vector space is unitary for some choice of inner product? Let $V$ a finite dimensional vector space over $\mathbb{C}$. Let $T\in GL(V)$.
Are there reasonable criteria for recognizing whether or not there is some inner product on $V$ w.r.t. to which $T$ is unitary? (equivalently, whether or not $T$ is similar to a unitary operator?)
 A: Sure.  An operator $T$ is unitary iff there is an orthonormal basis with respect to which $T$ is diagonal with eigenvalues of absolute value $1$.  So, $T\in GL(V)$ is unitary for some inner product iff it is diagonalizable with eigenvalues of absolute value $1$ (just pick an inner product which makes a basis of eigenvectors orthonormal).
A: Let $\langle \cdot, \cdot \rangle$ be the standard Hermitian inner product on $V \simeq \Bbb C^n$, and let $(\cdot, \cdot)$ be any other inner product; that is, $(\cdot, \cdot)$ is a (Hermitian) bilinear form on $V$, obeying as it must the rules
$(x, \alpha y) = \alpha(x, y) = (\bar \alpha x, y), \tag 1$
$(x, y + z) = (x, y) + (x, z), \tag 2$
$(x, y) = \overline{(y, x)}, \tag 3$
$x \ne 0 \Longrightarrow 0 < (x, x) \in \Bbb R, \tag 4$
$x = 0 \Longrightarrow (x, x) = 0; \tag 5$
we note that (2) and (3) yield
$(x + y, z) = \overline{(z, x + y)} = \overline{(z, x) + (z, y)} = \overline{(z, x)} +\overline{(z, y)}= (x, z) + (y,z); \tag 6$
we see then that any such $(\cdot, \cdot)$ is additive in both arguments, hence, bilinear.
Given $(\cdot, \cdot)$ and $x \in V$, we define a complex-linear functional $\phi(x): V \to \Bbb C$ by
$\phi(x):y \to (x, y); \tag 7$
it is well-known that such a functional satisfies
$\phi(x)(y) = (x, y) = \langle A(x), y \rangle \tag 8$
for some
$A(x) \in V. \tag 9$
It is also evident that $A$ itself is a linear map from $V$ to $V$, for
$\phi(x + z)(y) = (x + z, y) = \langle A(x + z), y \rangle, \tag{10}$
and also
$\phi(x + z)(y) = (x + z, y) = (x, y) + (z, y)$
$= \langle A(x), y \rangle + \langle A(z), y \rangle = \langle A(x) + A(z), y \rangle; \tag{11}$
since (10) and (11) hold for all $y \in V$ we must have
$A(x + z) = A(x) + A(z); \tag{12}$
likewise, 
$\langle A(\alpha x), y \rangle = (\alpha x, y) = \bar \alpha (x, y) = \bar \alpha \langle A(x), y \rangle = \langle \alpha A(x), y \rangle, \tag{13}$
whence
$A(\alpha x) = \alpha A(x); \tag{14}$
thus we see the linearity of
$A:V \to V. \tag{15}$
Now if $T$ is unitary with respect to $(\cdot, \cdot)$, we have
$\langle A(x), y \rangle = (x, y) = (Tx, Ty) = \langle A(T(x)), Ty \rangle = \langle T^\dagger AT(x), y \rangle, \tag{16}$
and thus
$T^\dagger AT = A; \tag{17}$
it should be remembered that $T^\dagger$ in this equation is the standard Hermitian adjoint of $T$ defined with respect to the standard inner product $\langle \cdot, \cdot \rangle$ on $V$; we also observe that if these two inner products on $V$ are in fact the same, so that indeed
$A = I, \tag{18}$
then (17) becomes
$T^\dagger T = I, \tag{19}$
and $T$ is an ordinary unitary operator on $V$.  
It is now a simple matter to walk these steps back and conclude that if $A$ satisfies (17), and is non-singular, then $T$ is unitary with repect to the inner product 
$(x, y) = \langle A(x), y \rangle. \tag{20}$
We note that (17) is linear in $A$, and thus there is no difficulty in principle in finding a solution.
