My contention is that there is an observable other than the identity operator, which is deterministic (meaning single eigenvalue). It is given as follows.
The Hilbert transform acts on functions of the form $f:\mathbb{R}\to\mathbb{R}$, to give out functions in the same domain, while preserving the norm. Let $f^h$ denote the Hilbert transform of $f$.
Let us stick to $1$-space dimension. Let all the wave functions live in a Hilbert space and be of unit norm.
We define the operator $i\mathcal{H}$ as follows. Let $\psi(x) = \psi_R(x) + i\psi_I(x)$, where $\psi_R, \psi_I$ are real avlued functions. Lets assume $||\psi|| = 1$. The operator $i\mathcal{H}$ is defined as follows.
$$i\mathcal{H}\psi = i(\psi_R^h + i\psi_I^h)$$
Properties of this operator $i\mathcal{H}$,
Linear and self-adjoint, and hence an observable.
Single eigenvalue $\lambda= 1$. The corresponding eigenspace consists of all the unit norm wave functions of the form $\psi = f+if^h$, where $f:\mathbb{R}\to \mathbb{R}$ and $||f|| = 0.5$
Edit : After comment from @Phoenix87, it has another eigenvalue $\lambda = -1$ and the corresponding eigen space is wave functions of the form $f-if^h$.
Note : $\lambda = 1$ corresponds to particle moving in positive $x$-direction, and $\lambda = -1$ corresponds to particle moving in negative $x$-direction. Wonder what I should name this operator!
The final state after the measurement is $$\frac{1}{2}{(\psi_R + \psi_I^h) +/- i\frac{1}{2}(\psi_I - \psi_R^h)}$$
Commutes with operators like translations in space and also with the momentum operator.
There may be more properties but these are the ones I could think of.
PS : The properties 1,2,3,4 can be verified using the fact that the Hilbert transform of $f^h = -f$ and few other properties of the Hilbert transform.