In a single particle Harmonic Oscillator, suppose I prepare it in the ground state and then measure its position. From the relation connecting Total Energy, Kinetic energy and Potential I can calculate its momentum also. So there is a one-to-one mapping between momentum and position. Then how can they have different uncertainties?
$(p^2/2m) + kx^2 = E_0$
When you measure the system's position you actually push it out of the ground state, so your "trick" of using the initial total energy to compute the momentum after the position measurement doesn't work.
To be specific, when you measure the position, suppose you get $x_0$ as the result. Then the particle's wave function is now a reasonably sharp peak at $x=x_0$. That is most certainly not the Gaussian wave function which you had when the system was in the ground state prior to the measurement. This peaked state actually doesn't even have a well defined energy. It is a quantum superposition of many different definite-energy states.
Before the measurement, the system is in the ground state $|0\rangle$. The wave function for this state expressed in the position basis is
$$\Psi(x) = \frac{1}{\sqrt{2\pi \sigma_x^2}} \exp [ -x^2 / (2\sigma_x^2) ] .$$
Note that this wavefunction does not have a single well-defined position. The value of $\sigma_x$ is in general not zero, so this is a Gaussian wave form with finite width.
Similarly, the wavefunction expressed in momentum basis is
$$\Psi(p) = \frac{1}{\sqrt{2\pi\sigma_p^2}} \exp[-p^2 / (2\sigma_p^2)] $$
where
$$\sigma_p = \frac{\hbar}{2} \frac{1}{\sigma_x} . $$
So again we see that the wave function in momentum has no specific single momentum. Note that this relationship between $\sigma_x$ and $\sigma_p$ is the tightest one allowed under the uncertainty relationship.
Now suppose we measure the position. When we do this we change the state of the system. Suppose we measure a value $x_m$ but our meter is such that it has an error of $\pm \delta x / 2$. Then the wave function after the measurement is something like
$$ \Psi'(x) = \begin{cases} 1/\delta x & x_m -\delta x / 2 < x < x_m + \delta x / 2 \\ 0 & \text{otherwise} \end{cases} $$
In other words, the wave function is now strongly localized near $x=x_m$. Note that this state is not an eigenstate of the energy (Hamiltonian operator).
Now OP suggests using the measurement of position, and the pre-measurement value of the energy to compute the momentum $p_i$ the system had prior to the measurement. Let $E_i$ denote the energy prior to the measurement, i.e. the ground state energy of the oscillator. Then OP argues something like this
$$ \begin{align} E_i &= \text{kinetic} + \text{potential} \qquad \qquad (1) \\ E_i &= p_i^2 / 2m + (1/2)k x_m^2 \qquad \qquad (2) \\ p_i &= \sqrt{2m[E_i - (1/2)k x_m^2]}. \qquad \qquad (3) \end{align} $$
Our job now is to explain what's wrong with this. The energy before and after the measurement are different. When we measure the particle and find it at $x_m$ the system now has a completely different energy than it did before the measurement. In fact, the state after the measurement is not an energy eigenstate, so it doesn't even have a specific energy. This invalidates Eq. $(2)$ because there we related the pre-measurement energy to the post-measurement position.
There are further problems with OP's calculation, but let's stop here and make sure we're all on the same page. If someone needs more clarification post a comment.
