How does the Lagrangian transform when coordinates are changed? I'll talk of single particle Lagrangian in $n$ dimensions.
Suppose in a given coordinate system with the coordinates $(q_i)_{i=1}^n$, the Lagrangian is given by $L(\mathbb{q,  \dot q}, t)$. Suppose I switch to another reference frame, which has coordinates $(q'_i)_{i=1}^n$. Now the Lagrangian for the primed system, $L'(\mathbb q', \mathbb{\dot q'}, t)$ must be such that $\mathbb q(t)$ is an E-L trajectory$^1$ in the primed frame if and only if $\mathbb q'(t):=\mathbb f (\mathbb q(t), t)$ is an E-L trajectory in the unprimed frame, where $\mathbb f$ is the function from $\mathbb R^{n+1}$ to $\mathbb R^n$ that gives the coordinate transformation from the unprimed space to the primed space. (Note: Having $t$-dependence in $\mathbb f$ allows to consider frames "moving" w.r.t. each other also.)
I suspect that the relation between the Lagrangians in the two frames will be given by$^2$
$$
L(\mathbb q, \mathbb{\dot q}, t) = L'\left(\mathbb{f(q,} t), \mathbb{\frac{\partial f}{\partial q}} (\mathbb q, t)\cdot\mathbb{\dot q} + \frac{\partial \mathbb f}{\partial t}(\mathbb q, t), t\right).
$$
My work so far:
I have been able to show that for any E-L trajectory $\mathbb q(t)$ in the unprimed frame, for the above mentioned $L'$, we have$^3$
$$
\left[\frac{\partial L'}{\partial \mathbb q'} (\circ) - \frac{\mathrm d}{\mathrm dt} \frac{\partial L'}{\partial\mathbb{\dot q'}} (\circ)\right]\cdot\frac{\partial\mathbb f}{\partial\mathbb q}(\mathbb q(t), t) = 0,
$$
where $\circ:=\left(\mathbb f(\mathbb q(t)), t), \mathbb{\frac{\partial f}{\partial q}} (\mathbb q(t), t)\cdot\mathbb{\dot q}(t) + \frac{\partial \mathbb f}{\partial t}(\mathbb q(t), t), t\right)$. Bow I'm not able to conclude that
$$
\frac{\partial L'}{\partial \mathbb q'} (\circ) - \frac{\mathrm d}{\mathrm dt} \frac{\partial L'}{\partial\mathbb{\dot q'}} (\circ) =0,
$$
which is what I aim to prove in order to prove my assertion.
My further thoughts: I believe that I will need to use "invertibility" of $\mathbb f$ since I've not used that so far.

$^1$A trajectory which satisfies E-L equations.
$^2$$\mathbb{\frac{\partial f}{\partial q}} (\mathbb q, t)\cdot\mathbb{\dot q}$ is a vector whose $i$th component is given by $\sum_j\frac{\partial f_i}{\partial q_j}(\mathbb q, t)\dot q_j$
$^3$Tell me if clarification in "vector differentiation" here is needed.
 A: $$\det\left[\frac{df_h}{dq_k}\right] \neq 0 \tag{1}$$
because we are dealing with a bijective and bi-differentiable  transformation of coordinates (see below). Hence, if
$$
\left[\frac{\partial L'}{\partial \mathbb q'} (\circ) - \frac{\mathrm d}{\mathrm dt} \frac{\partial L'}{\partial\mathbb{\dot q'}} (\circ)\right]\cdot\frac{\partial\mathbb f}{\partial\mathbb q}(\mathbb q(t), t) = 0,
$$
then, by multiplying both sides with $\left(\frac{\partial\mathbb f}{\partial\mathbb q}\right)^{-1}$ that exists due to (1), we conclude that the only possibility is
$$
\frac{\partial L'}{\partial \mathbb q'} (\circ) - \frac{\mathrm d}{\mathrm dt} \frac{\partial L'}{\partial\mathbb{\dot q'}} (\circ) =0\:.
$$
Namely $t \mapsto \mathbb{f}(\mathbb{q}(t),t)$ solves the EL equations with rrspect to $L'$ if (and only if) $t \mapsto \mathbb{q}(t)$ does with respect to $L$.
I finally prove that (1) is true. We are here dealing with two coordinate sistems $t, q_1; \ldots, q_n$ and $t', q'_1; \ldots, q'_n$ and we are assuming that
$$t'=t\:, \quad q'_k = f_k(q_1,\ldots, q_n, t)$$
where this trasformation is differentiable, bijective with differentiable inverse as is requested for local charts on the same manifold. Hence we can write in particular, at fixed $t$,
$$q'_k = q'_k(q_1(q'_1,\ldots, q'_n,t),\ldots, q_n(q'_1,\ldots, q'_n,t), t)$$ so that
$$\delta_{kr} = \frac{\partial q'_k}{\partial q'_r}
= \sum_{j=1}^n \frac{\partial q'_k}{\partial q_j} \frac{\partial q_j}{\partial q'_r}$$
In other words
$$I = \left[ \frac{\partial q'_k}{\partial q_j} \right]_{k,j=1,\ldots, n}
\left[ \frac{\partial q_j}{\partial q'_r} \right]_{j,r=1,\ldots, n}\:.$$
We have established that the Jacobian matrix $$\frac{\partial \mathbb{f}}{\partial \mathbb{q}}= \left[ \frac{\partial q'_k}{\partial q_j} \right]_{k,j=1,\ldots, n}$$
is in particular bijective so that (1) is valid.
