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Let $V$ a finite dimensional vector space over $mathbb{C}$. Let $Tin GL(V)$.

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?)

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, $Tin GL(V)$ 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).

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.