My question is that can I say for every specified LTI system, the
impulse response function is fixed due to the nature of this system?
This is bit subtle but a non-LTI system does not have an impulse response \$h(t)\$ (or an \$h[n]\$ for discrete time systems).
If the system is linear but not time invariant and the input is a delayed impulse \$\delta(t - \tau)\$, the output \$h(t,\tau)\$ is a function of both \$t\$ and \$\tau\$.
For example, consider the system defined by
$$y(t) = u(t)x(t)$$
For \$t \lt 0\$, the system output is zero regardless of the input. For \$t \gt 0\$, the system output is identical to the input.
Clearly, this system is not time invariant. If the input is \$x(t - \tau)\$, the output is
$$y(t, \tau) = u(t)x(t - \tau)$$
but
$$y(t - \tau) = u(t - \tau)x(t - \tau) \ne y(t,\tau)$$
so the output due to a delayed input is not equal to the delayed version of the output from a non-delayed input.
We can't meaningfully speak of an impulse response for this system since, for an impulse before \$t = 0\$, the output is zero whilst for an impulse after \$t = 0\$, the output is an impulse.
Likewise, we can't meaningfully speak of a frequency response for this time variant system.
The Fourier transform of the system is
$$Y(j \omega) = \frac{1}{j\omega} \ast X(j \omega) + \pi X(j\omega)$$
which is clearly not of the form
$$Y(j \omega) = H(j \omega) X(j \omega)$$
as required for there to be a transfer function (or frequency response) for the system.
All of the above is simply to show that an LTI system has an impulse response and linear time-variant systems do not.