Short Answer
The harmonic amplitudes of the longitudinal "pluck" would follow the same pattern as the transverse pluck.
Longer Answer
Without having to rederive the harmonic amplitudes for the new case, we can just compare the relevant wave equations and boundary/initial conditions of the new problem and see if they are of the same form as the transverse case. If the setups are of the same form, then the harmonic amplitudes should also be of the same form.
Transverse Problem
The small-amplitude wave equation for transverse excitation of a string may be written as
$$ \frac{\partial^2 y}{\partial x^2} = \frac{\mu}{T}\frac{\partial^2y}{\partial t^2}, $$
where $y$ is the transverse displacement of the string, $x$ is the position along the string, $\mu$ is the linear mass density of the string, and $T$ is the tension in the string. This second-order partial differential equation is known as the wave equation and requires two boundary conditions and two initial conditions to solve. The boundary conditions are that the string is fixed at the endpoints (e.g., $y=0$ at the edges), and the initial conditions are that the initial velocity is zero (the string is released from rest) and the initial displacement follows the pattern of a plucked string (a triangular shape). We can express these conditions mathematically as
\begin{gather}
y(0,t)=y(L,t)=0, \\
\left.\frac{\partial y}{\partial t}\right|_{x,t=0}=0, \\
y(x,0) = f(x),
\end{gather}
where the endpoints are given at $x=0$ and $x=L$ and $f(x)$ represents the triangular function of the pre-plucked string.
Longitudinal Problem
The longitudinal problem is the elastic bar problem, and derivations are rather plentiful (see this one, for instance). While not strictly necessary to answer this question, I have decided to provide a hopefully-intuitive derivation below. Hopefully the derivation provides a richer explanation of the answer at hand.
Imagine a system made up of many small masses that are connected to each other by springs. In the limit as the masses approach infinitesimal in size and mass and infinite in number we can approximate the system as the "ideal" string in question. We can use this model to derive a dynamic equation for longitudinal waves in the system. Consider the $i$th mass with mass $dm=\mu dx$ (where $dx$ is the spacing between masses) with identical springs with spring constant $k$ on either side. Then the net force on this mass may be written as
$$ F_i = -k(u_i-u_{i-1}) - k(u_i - u_{i+1}) = k(u_{i+1}-2u_i+u_{i-1}), $$
where $u_i$ is the displacement of the $i$th mass (i.e., the current position minus the rest position). Then, by Newton's second law we may write
$$ \mu dx\frac{\partial^2u_i}{\partial t^2} = k(u_{i+1}-2u_i+u_{i-1}). $$
The displacement of the different masses may be thought of as various evaluations of a displacement field (called the particle displacement), such that we may write
\begin{gather}
u_i(t) = u(x_i,t), \\
u_{i+1}(t) = u(x_i+dx,t), \\
u_{i-1}(t) = u(x_i-dx,t).
\end{gather}
(Note that we are assuming that the displacement is small compared to $dx$, and so this analysis is the small signal approximation.) If we assume that $u$ is differentiable, then as $dx\rightarrow0$ we may obtain
\begin{gather}
u_{i+1}(t) = u(x_i,t) + dx\left.\frac{\partial u}{\partial x_i}\right|_{x_i,t} + \frac{dx^2}{2}\left.\frac{\partial^2 u}{\partial x_i^2}\right|_{x_i,t} + O(dx^3), \\
u_{i-1}(t) = u(x_i,t) - dx\left.\frac{\partial u}{\partial x_i}\right|_{x_i,t} + \frac{dx^2}{2}\left.\frac{\partial^2 u}{\partial x_i^2}\right|_{x_i,t} + O(dx^3).
\end{gather}
Substituting these expressions into Newton's second law, rearranging, and exchanging the variable $x$ for $x_i$, we then obtain the wave equation
$$ \frac{\partial^2 u}{\partial x^2} = \frac{\mu}{k\ dx}\frac{\partial^2 u}{\partial t^2}, $$
where $k\ dx$ represents an elastic modulus, specifically the Young's modulus multiplied by the cross-sectional area.
At this point we need to define the boundary conditions at $x=0$ and $x=L$. If we assume that the connection sites are rigid and clamped, then we require
\begin{gather}
u(0,t)=u(L,t)=0.
\end{gather}
Then, we "pluck" the string by forcing its displacement profile into the shape of $f(x)$ and letting go, or
\begin{gather}
u(x,0) = f(x), \\
\left.\frac{\partial u}{\partial t}\right|_{x,0} = 0.
\end{gather}
But these conditions are just the same as the conditions for the transverse vibration of the string! In fact, the only difference between the two questions from a mathematical perspective is the wave speed. Then, noting that the mode shapes do not depend on the wave speed, we must conclude that the two systems share the same mode shapes, and the conclusion obtained in the other post holds.
Final Notes
The resonance frequencies of the longitudinal vibrations are very different from the resonance frequencies of the transverse vibrations, as they depend on the wave speed. Thus, details of the amplitudes for the two systems will vary, even if they both are started with the same "pluck" function $f(x)$.
I have also emphasized that the equations above are for small signals. Indeed, to be perfectly valid the signals need to be infinitesimal in size. As the signals become larger, additional effects become present, and the transverse and longitudinal modes will become coupled. The real world is never a clean and pretty as our math would like it to be. :)
