If $\langle v, Av \rangle = 1$ for all $\|v\| = 1$, then $A = I$. Theorem: If $\langle v, Av \rangle = 1$ for all $\|v\| = 1$, then $A = I$, where $A$ is a bounded operator from $\mathbb{C}^n$ to $\mathbb{C}^n$.
From what I can tell, converting to the form of $v^\dagger A^\dagger v$, can obtain that the diagonal elements of the matrix form of $A$ must all be one, by subbing in each basis element for $v$ in the equation, but I cannot prove the rest (that the rest of the matrix is zeroes). Maybe another method is correct. I think maybe something to do with eigendecomposition might be the right direction (show A must decompose into eigenvalues all 1 with identity basis).
 A: This is another approach.

Lemma: if a matrix $B$ satisfies $\langle v, Bv \rangle = 0$ for all $v$, then $B = 0$.
Proof: For any $x$ and $y$, we have:
$$\begin{array}{rcl}
\langle x+y, B(x+y) \rangle &=& 0 \\
\langle x, Bx \rangle + \langle x, By \rangle + \langle y, Bx \rangle + \langle y, By \rangle &=& 0 \\
0 + \langle x, By \rangle + \langle y, Bx \rangle + 0 &=& 0 \\
\langle x, By \rangle &=& -\langle y, Bx \rangle
\end{array}$$
Therefore, $B$ is skew-symmetric, i.e. $B = -B^*$.
Then, we have $BB^* = B^*B$, since $BB^* = -B^*B^* = B^*B$.
From the spectral theorem, $B$ is diagonalizable, i.e. it has $n$ linearly independent eigenvectors. However, the eigenvalues must all be $0$.
Therefore, the eigenspace of $0$ of $B$ is the full space, so $B$ is $0$.

Consider $B = A-I$. Then, for every $v$, $\langle v,Bv \rangle = \langle v,Av \rangle - \langle v,v \rangle = 0$.
Therefore, from the lemma above, $B=0$, i.e. $A=I$.
A: $\newcommand{\la}{\langle}\newcommand{\ra}{\rangle}
\newcommand{\id}{\operatorname{id}}$
\begin{align*}
& \langle v, Av\rangle=\la v,\id v\ra \forall \| v\|=1\\
\Longrightarrow &  \langle v, Av\rangle=\la v,\id v\ra \forall v \\
\Longrightarrow & \langle u, Av\rangle=\la u,\id v\ra \forall u,v (\mbox{ by polarization } \la u,Av\ra =\sum_{k=0}^3 i^k \la u+i^k v, A(u+i^k v)\ra )\\
\Longrightarrow & A=\id .
\end{align*}
A: Your condition is equivalent to $\langle v, Av \rangle = \|v\|^2$ for all $v$.

I now prove that there are $n$ mutually orthogonal eigenvectors with eigenvalue $1$:
By the fundamental theorem of algebra, there is an eigenpair $(\lambda,v)$ of $A$. Then, $\overline \lambda \|v\|^2 = \|v\|^2$ by the condition and by properties of complex inner product, whence $\lambda = 1$.
Now, prove that the orthogonal complement of $v$ also satisfies the condition, and proceed by induction (since the orthogonal complement has one dimension lower than the original vector space).

Then, the eigenspace is the whole space, and the eigenvalue is $1$, so the matrix must be the identity matrix.
A: A way to see this at a glance:

If $X$ is a Hilbert space and  $A : X \to X$ is a bounded operator such that $\langle Ax,x\rangle \in \mathbb{R}$ for all $x \in X$ then $A$ is self-adjoint.

$\langle Av, v\rangle = \|v\|^2 \in \mathbb{R}$ for all $v \in \mathbb{C}^n$ so $A$ is self-adjoint. In particular $A$ is diagonalizable.
The condition $\langle Av, v\rangle = \|v\|^2$ also gives that all eigenvalues of $A$ are equal to $1$:
$$\lambda \|v\|^2 = \langle \lambda v, v\rangle = \langle Av, v\rangle = \|v\|^2 \implies \lambda = 1$$
Hence $A$ diagonalizes to the identity matrix, so $A = I$.
