Under what conditions is a linear automorphism an isometry of some inner product? Assume $V$ is a finite-dimensional vector space over $\mathbb{R}$, and that $T: V \to V$ is a (linear) isomorphism. 

When is it possible to construct an inner product on $V$ 
  making $T$ an isometry?

(Hopefully, I am looking for necessary & sufficient conditions $T$ should satisfy, i.e. a full characterization of the situation).
What I have so far:
A necessary condition: all the real eigenvalues of $T$ are of absolute value $1$. (Since $ T(v)=\lambda v \Rightarrow  \langle v,v\rangle=\langle Tv,Tv\rangle = \langle \lambda v, \lambda v\rangle = \lambda^2\langle v, v\rangle$ and an eigenvector $v$ must be nonzero.)
This condition is certainly not sufficient:
For example look at $A$ = $\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}: \mathbb{R}^2 \to \mathbb{R}^2$. It is an automorphism which has only one eigenvalue ($\lambda = 1$). However, $A\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+y \\ y \end{pmatrix}$, hence $A^n\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+ny \\ y \end{pmatrix}$ and the requirement $A:(\mathbb{R}^2,\langle \rangle) \to (\mathbb{R}^2,\langle \rangle) $ to be an isometry for some inner product $\langle \rangle$ implies: $\lVert \begin{pmatrix} x \\ y \end{pmatrix}\rVert^2=\lVert A^n\begin{pmatrix} x \\ y \end{pmatrix}\rVert^2\Rightarrow x^2 \lVert e_1\rVert^2+y^2 \lVert e_2\rVert^2+2xy\langle e_1,e_2\rangle = (x+ny)^2 \lVert e_1\rVert^2+y^2 \lVert e_2\rVert^2+2y(x+ny) \langle e_1,e_2\rangle \Rightarrow 0=(2nxy+n^2y^2)\lVert e_1\rVert^2+2ny^2 \langle e_1,e_2\rangle$. 
So we get that $0=(2xy+ny^2)\lVert e_1\rVert^2+2y^2 \langle e_1,e_2\rangle$ for any $x,y\in \mathbb{R}, n\in \mathbb{N}$ which is a contradiction since $\lVert e_1 \rVert > 0$.
Some sufficient conditions:
1)  If $T$ is diagonalizable over $\mathbb {R}$ (with all eigenvalues $1$ or $-1$, by our necessary condition), then let ${V_1,...,V_n}$ be a basis of eigenvectors of $T$ , and define $\langle v_i,v_j\rangle = \delta_{ij}$. $T$ will be an isometry.
This condition is certainly not necessary:
just take a rotation (say by $90^{\circ}$) in the plane. note that it is diagonalizable over $\mathbb{C}$. My guess is that if our transformation is diagonalizable over $\mathbb{C}$ (with all eigenvalues with absolute value 1) a similar construction like the above will work. One problem I see with this approach is that an odd-dimensional $\mathbb{R}$-vector space cannot even be considered as a $\mathbb{C}$-vector space. (Though we can always complexify...).
2) $T$ is of finite order. (Then we just start with any inner product on $V$ and construct a new one via summing over iterates of $T$, i.e: $\langle v,w \rangle ' = \sum_{i=0}^{n-1} \langle T^iv,T^iw \rangle $). Note that (as explained for instance here) this implies $T$ is diagonalizable over $\mathbb{C}$, but of course not necessarily over $\mathbb{R}$. (Think about our rotation again.)
Actually, I have now understood that condition (1) implies $T$ is of order 2, (I think the reverse implication also holds, i.e $T^2=Id\Rightarrow T$ is diagonalizable). So condition (1) is a particular case of (2).
However, (2) is  not necessary, since any rotation of irrational multiple of 2$\pi$ is an isometry w.r.t the standard product, but of infinite order.
I somehow think the right way to handle this question is to think over $\mathbb{C}$, but I am not sure how to do this.
 A: Hint If $T$ is an isometry of the inner product $(x, y) \mapsto \langle x, y \rangle$, then for any $P \in GL(V)$, $P^{-1} T P$ is an isometry of the inner product $(x, y) \mapsto \langle P x, P y \rangle$ (it is not hard to verify that this indeed defines an inner product, but you may wish to prove it anyway): Thus, the property that a given transformation admit such an inner product is invariant under similarity (that is, is an invariant of the conjugacy class of $T$ in $GL(V)$).
On the other hand, we already know canonical representatives of each similar class in $GL(V)$: These are given by the analogue of the Jordan normal form for real matrices. These matrices are relatively simple and so one can check more directly for a general matrix of this form whether such an inner product exists.
Instead of attacking this immediately, you might like to prove first the following lemmata:


*

*One can essentially treat each "Jordan block" separately, or more precisely, given a direct sum decomposition $V = \oplus V_a$ and linear transformations $T_a : V_a \to V_a$, then there is an inner product on $V$ preserved by $T := \oplus T_a$ iff for each $a$ there is an inner product on $V_a$ preserved by $T_a$.

*$T$ cannot have any nonsimple "Jordan blocks", that is, $T$ is block diagonal, where each block has the form $$\phantom{(\ast)} \qquad \begin{pmatrix} \lambda \end{pmatrix} \qquad \text{or} \qquad \begin{pmatrix} \alpha & -\beta \\ \beta & \alpha\end{pmatrix} \qquad (\ast).$$
The proof of (2) is essentially the one you give for your counterexample $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.$
With these two facts in hand, it's essentially enough to solve the problem for the two transformations in $(\ast)$ above. Since the sole eigenvalue of $\begin{pmatrix} \lambda \end{pmatrix}$ is $\lambda$, by the first observation in the question a sufficient and necessary condition for this block is $\lambda \in \{\pm 1\}$.
A: Curious that the simple necessary and sufficient condition has not been (it would seem to me) clearly mentioned in any of the answers so far:

The linear operator $T$ preserves some inner product on $V$ if and only if $V$ admits a basis for which the matrix of $T$ is orthogonal (in other words the matrix of $T$ on an arbitrary basis is similar to an orthogonal matrix). The occurs if and only if the complexification of $T$ is diagonalisable, and all its (complex) eigenvalues have absolute value$~1$.

For the "if" of the first sentence, it suffices to define the inner product by the formula for the standard inner product of $\Bbb R^n$ in terms of the coordinates with respect to that basis; for the converse (only if) just take any orthonormal basis for the inner product. That the conditions in the second sentence are necessary is because these are well-known properties of orthogonal matrices.
That they are sufficient, in other words that a real matrix$~A$ that is diagonalisable over $\Bbb C$ with all its eigenvalues of absolute value$~1$ is similar (over$~\Bbb R$) to an orthogonal matrix is also standard, but the argument is slightly more involved. Since the complex-linear operator$~\phi$ on$~\Bbb C^n$ defined by$~A$ commutes with the (real-linear) operation$~J$ that performs complex conjugation of all coordinates, one easily shows that a basis of eigenvectors for$~\phi$ can be chosen such that as a set it is stable under$~J$: eigenvectors for real eigenvalues are chosen to be $J$-fixed (have real coordinates), and eigenvectors in the basis for non-real eigenvalues to come in pairs interchanged by$~J$, whose eigenvalues are complex conjugates of each other. This gives a decomposition of$~\Bbb C^n$ into a direct sum of $J$-stable subspaces of (complex) dimension $1$ or$~2$, each of which is the complexification of its real subspace of $J$-fixed vectors. These define a decomposition of (the original real vectors space) $V$ into $T$-stable subspaces of (real) dimension $1$ or$~2$, and it suffices to show that each of these subspaces has a basis for which the matrix of the restriction of$~T$ to the subspace is orthogonal. In dimension$~1$ this is trivial (the restriction is scalar multiplication by $1$ or by$~{-}1$) while in dimension$~2$ one can take a basis consisting of the common "real part" of the pair of complex eigenvectors and of the common (up to sign) "imaginary part", for which the $2\times2$ matrix will be that of a rotation, hence orthogonal.
A: Here is another answer which I think in some sense is much better, in some not. 
Summary: we reduce to the usual case of finding a 'nice' form for orthogonal matrices.
Assume there exists an inner product $\langle \rangle$ on $V$  making $T$ an isometry. Then there exists an orthonormal basis $B=(v_1,...,v_n)$. Now look at the representing matrix of our automorphism $T$ w.r.t to $B$: $A=[T]_B$. Then $[Tv]_B=[T]_B[v]_B$, where $[u]_B$ is the coordiante vector of $u\in V$ w.r.t $B$. 
It clearly holds $\langle Tv_i,Tv_j \rangle = \langle v_i,v_j \rangle= \delta_{ij}$. Now note that bilinearity of the inner product implies: 
$\langle Tv_i,Tv_j \rangle = \langle [Tv_i]_B,[Tv_j]_B \rangle_{Euclidean}= \langle [T]_B[v_i]_B,[T]_B[v_j]_B \rangle_{Euclidean} = \langle Ae_i,Ae_j \rangle_{Euclidean} = \langle A_{i\downarrow} ,A_{j\downarrow} \rangle_{Euclidean}$.
So finally we get: $\langle A_{i\downarrow} ,A_{j\downarrow} \rangle_{Euclidean}=\delta_{ij}$, so the columns of $A$ form an orthonormal basis for $\mathbb{R}^n$, hence $A=[T]_B$ is an orthogonal matrix. 
Now by the real canonical form of an orthogonal matrix $A$ is similar over $\mathbb{R}$ to a matrix with simple real jordan blocks as required, or equivalently diagonalizable over $\mathbb{C}$. That means there exists a basis that w.r.t to it $T$ has the canonical form. 
I would still like to find a more geometric\conceptual reaosn for why non simple jordan blocks can never preserve an inner product. (My direct proof was quite computational). Ofcourse we can resort to the uniqueness argument of Jordan form, (We showed any admissible transformation has a simple Jordan form, and the Jordan form is unique up to the order of the blocks, and that's it).
A: For sake of completeness I write here in detail part of the solution, following the suggestions of Travis. 
By lemma1 it is enough to check the case of a single Jordan Block.
First assume  $A$ = $\begin{pmatrix} \lambda & 1 \\\ 0 & \lambda \end{pmatrix} $ is the matrix representation of our operator $T : V \to V$ w.r.t some basis $v_1,v_2$ of $V$. 
Hence $Av_1 = \lambda v_1$, $Av_2 = v_1+\lambda v_2$ , so  $T^n (xv_1+yv_2)= x\lambda^nv_1+y(n\lambda^{n-1}v_1+\lambda^nv_2)= (\lambda^nx+n\lambda^{n-1}y)v_1+ (\lambda^ny)v_2$. 
In coordinates-vector rep: $A^n\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} \lambda^nx+n\lambda^{n-1}y \\ \lambda^ny \end{pmatrix}$ (can be verified easily via induction).
Now, the requirement for $T:(V,\langle \rangle) \to (V,\langle \rangle) $ to be an isometry for some inner product $\langle \rangle$ implies: 
$ \lVert T^n (xv_1+yv_2) \rVert^2=\lVert (xv_1+yv_2) \rVert^2$ , i.e
$\lVert \begin{pmatrix} x \\ y \end{pmatrix}\rVert^2=\lVert A^n\begin{pmatrix} x \\ y \end{pmatrix}\rVert^2\Rightarrow x^2 \lVert e_1\rVert^2+y^2 \lVert e_2\rVert^2+2xy\langle e_1,e_2\rangle = (\lambda^nx+n\lambda^{n-1}y)^2 \lVert e_1\rVert^2+\lambda^{2n}y^2 \lVert e_2\rVert^2+2(\lambda^ny)(\lambda^nx+n\lambda^{n-1}y) \langle e_1,e_2\rangle $
$\Rightarrow $ (using the fact that $\lambda\in$ {$-1,1$}, so $\lambda^{2n}=1$, we get:
$ 0=(2nx\lambda^{2n-1} y+n^2y^2\lambda^{2n-2})\lVert e_1\rVert^2+2ny^2\lambda^{2n-1} \langle e_1,e_2\rangle$. 
So finally (after dividing by $n\lambda^{2n-2}$ ) we attain:
$0=(2xy\lambda+ny^2)\lVert e_1\rVert^2+2y^2\lambda \langle e_1,e_2\rangle$ for any $x,y\in \mathbb{R}, n\in \mathbb{N}$ which is a contradiction since $\lVert e_1 \rVert > 0$.
A similar (but more tedious...) computation shows a similar contradiction will happen if look at larger (size $>2$) blocks.
The case of a "Real" Jordan block can also be treated analogously.
What is left to check is what kind of simple blocks can appear?
The case for $(\lambda)$ is indeed trivial according to my first observation.
We now state and prove the second case.
Claim: $A = \begin{pmatrix} a & b \\\ {-b} & a \end{pmatrix} $ represents (w.r.t some basis) a suitable operator $T$ (one which admits an inner prodcut such that...) if and only if $|a^2+b^2|=1$
Proof:
First assume  $A = \begin{pmatrix} a & b \\\ {-b} & a \end{pmatrix} $ is the matrix representation of our operator $T : V \to V$ w.r.t some basis $v_1,v_2$ of $V$. Assume also that $\langle ,\rangle $ is an inner product on $V$ w.r.t $T$ is an isometry. Then
$\lVert \begin{pmatrix} x \\ y \end{pmatrix}\rVert^2=\lVert A\begin{pmatrix} x \\ y \end{pmatrix}\rVert^2 = \lVert \begin{pmatrix} ax+by \\ -bx+ay \end{pmatrix}\rVert^2 \Rightarrow x^2 \lVert e_1\rVert^2+y^2 \lVert e_2\rVert^2+2xy\langle e_1,e_2\rangle = (ax+by)^2 \lVert e_1\rVert^2+(-bx+ay)^2 \lVert e_2\rVert^2+2(ax+by)(-bx+ay) \langle e_1,e_2\rangle $
Now $y=0 \Rightarrow x^2\lVert e_1\rVert^2=a^2x^2\lVert e_1\rVert^2+ b^2x^2\lVert e_2\rVert^2-2abx^2\langle e_1,e_2\rangle $
$\Rightarrow (*) \lVert e_1\rVert^2=a^2\lVert e_1\rVert^2+ b^2\lVert e_2\rVert^2-2ab\langle e_1,e_2\rangle$ 
$x=0$ $\Rightarrow y^2\lVert e_2\rVert^2=b^2y^2\lVert e_1\rVert^2+ a^2y^2\lVert e_2\rVert^2+2aby^2\langle e_1,e_2\rangle $ 
$\Rightarrow (**) \lVert e_2\rVert^2=b^2\lVert e_1\rVert^2+ a^2\lVert e_2\rVert^2+2ab\langle e_1,e_2\rangle$ 
Now we add $(*),(**)$ and we get:
$\lVert e_1\rVert^2+ \lVert e_2\rVert^2=(a^2+b^2)(\lVert e_1\rVert^2+ \lVert e_2\rVert^2)$ 
This forces $|a^2+b^2|=1$ as required. 
Now for the other direction:
Let  $A = \begin{pmatrix} a & b \\\ {-b} & a \end{pmatrix} $ be the matrix representation of $T : V \to V$ w.r.t some basis $e_1,e_2$ of $V$. Assume $|a^2+b^2|=1$. Now define $\langle e_1,e_2\rangle = 0 ,\lVert e_1\rVert^2 = \lVert e_2\rVert^2 = 1 $. It is now a straightforward check to verify $T$ is an isometry w.r.t this inner product. (it takes the orthonormal basis $(e_1,e_2)$ to an orthonormal basis).
I must admit I am still not completely satisfied by this sufficient & necessary condition, I was hoping for something more "clean and simple". But now I think I see some of the inherent complexity in the question. (There cannot be a characterization based on the identity of the eigenvalues alone, since Id and my counterexample share these).
