Since it's part of the definition of an inner product space that it must be over the real or complex numbers, any vector space over some other field will not admit an inner product.
Barring that obvious obstruction, the situation where one's vector space $V$ is over $\mathbb K$ ($=\mathbb R$ or $\mathbb C$) is quite simple: For any basis $v_1,v_2,\dotsc,v_n$ one has an inner product given by
$\qquad \displaystyle\langle v_i,v_j\rangle = \delta_{ij} = \begin{cases} 1 & \text{if }i=j \cr 0 & \text{if }i\neq j\end{cases}\ $ extended using sesquilinearity to all of $V$.
Some related things you might find enlightening:
Norms Induced by Inner Products and the Parallelogram Law: If a norm satisfies the parallelogram law, does it come from an inner product? Yes, but it takes a surprising amount of work to actually show that this is the case.
Any two norms on a finite dimensional linear space (vector space over $\mathbb R$ or $\mathbb C$) are equivalent. If you search Math.SE you're likely to find several questions and answers related to this. It's very common to prove this in real analysis courses.
Suppose $V$ is a finite dimensional normed linear space and $K \subset V$ is compact , absolutely convex and spans $V$, then there is a norm on $V$ such that $K$ is the unit ball w.r.t. that norm (and per the above equivalent to the given norm on $V$). It's not too difficult to come up with an idea that could workâjust draw a picture of the situation in $\mathbb R^2$.
On a finite-dimensional real inner product space, the notions of hermitian and symmetric for operators coincide; that is,
$T^\dagger = T^t = T; \tag 1$
since $T$ is symmetric, it may be diagonalized by some orthogonal operator
$O:V \to V, \tag 2$
$OO^t = O^tO = I, \tag 3$
$OTO^t = D = \text{diag} (t_1, t_2, \ldots, t_m), \; t_i \in \Bbb R, \; 1 \le i \le m, \tag 4$
where
$m = \dim_{\Bbb R}V; \tag 5$
since each $t_i \in \Bbb R$, for odd $n \in \Bbb N$ there exists
$\rho_i \in \Bbb R, \; \rho_i^n =t_i; \tag 6$
we observe that this assertion fails for even $n$, since negative reals do not have even roots; we set
$R = \text{diag} (\rho_1, \rho_2, \ldots, \rho_m); \tag 7$
then
$R^n = \text{diag} (\rho_1^n, \rho_2^n, \ldots, \rho_m^n) =
\text{diag} (t_1, t_2, \ldots, t_m) = D; \tag 8$
it follows that
$T = O^tDO = O^tR^nO = (O^tRO)^n, \tag 9$
where we have used the general property that matrix conjugation preserves products:
$O^tAOO^TBO = O^tABO, \tag{10}$
in affirming (9). We close by simply setting
$S = O^tRO. \tag{11}$
Best Answer
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).