Let the velocity of the transmitter be $\ \mathbf{v}\ $, the times at which it emits $\ T\ $ radio pulses be $\ t_1, t_2,\dots, t_T\ $, and its position at time $\ t=t_1\ $ be $\ \mathbf{x}_0\ $. All these quantities are unknown.
Let the positions of the $\ R\ $ receivers be $\ \mathbf{p}_r, r=1,2,\dots,R\ $ and the time when the $\ i^{\,\text{th}}\ $ pulse arrives at the $\ r^{\,\text{th}}\ $ receiver be $\ a_{ir}\ $. All these quantities are known.
The position of the transmitter at time $\ t\ $ will be $\ \mathbf{x}_0+\mathbf{v}\big(t-t_1\big)\ $, so the distance between the transmitter and the $\ r^{\,\text{th}}\ $ receiver at the time when it transmits the $\ i^{\,\text{th}}\ $ pulse will be $\ \big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|\ $. The time the pulse will take to travel this distance is $\ c^{-1}\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|\ $, therefore it will arrive at the $\ r^{\,\text{th}}\ $ receiver at time $\ t_i+$$c^{-1}\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|\ $, which we know to be $\ a_{ir}\ $. Therefore,
\begin{align}
(1)\hspace{1em}c\big(a_{ir}-t_i\big)&=\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|
\end{align}
for $\ i=1,2,\dots,T\ $ and $\ r=1,2,\dots,R\ $. Squaring this gives
\begin{align}
(2)\hspace{1em}c^2\big(a_{ir}-t_i\big)^2&=\big\|\mathbf{v}\big\|^2\big(t_i-t_1\big)^2+2\big(t_i-t_1\big)\big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle+\big\|\mathbf{x}_0-\mathbf{p}_r\big\|^2\ .
\end{align}
In these equations, we can take $\ t_i,\ $$i=1,2,\dots,T\ $, $\big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle,\ $$\big\|\mathbf{x}_0-\mathbf{p}_r\big\|^2,\ $$r=1,2,\dots,R\ $, and $\ \big\|\mathbf{v}\big\|^2\ $ as the unknown real numbers whose values we need to recover. There are a total of $\ T+2R+1\ $ of these, and we have $\ TR\ $ equations available to determine them, so if $\ TR\ge T+2R+1\ $, then there should only be a small finite number of solutions. If $\ R\ge2\ $, this will be true if and only if $\ T\ge\frac{2R+1}{R-1}\ $. For $\ R=2\ $, $\ T=5\ $ will be enough to achieve this, while for $\ R=3\ $, $\ T=4\ $ will suffice.
Even if one of these conditions is satisfied, there's unlikely to be a convenient formula for the solutions, but it should be possible to obtain
good approximations to them by a numerical algorithm, such as the Newton-Raphson method, for instance. When the system of equations is overdetermined $\big($that is, $\ TR>T+2R+1\ $, and all the equations are $\text{used}\big)$, it will, in practice, have no exact solution anyway, because imprecision in the measurements of $\ a_{ir}\ $ means that the values we have for them are only approximations. In this case, the Newton-Raphson method will find the values of the unknown quantities that minimise the sum of squares
$$
\sum_{i=1}^T\sum_{r=1}^R\Big(c^2\big(a_{ir}-t_i\big)^2-\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|^2\Big)^2\ .
$$
Once we have determined the values of all the unknown quantities, we can obtain the speed of the transmitter by taking the square root of $\ \|\mathbf{v}\|^2\ $, and we can try to use the now known values of $\ \|\mathbf{v}\|,$$\,\big\|\mathbf{x}_0-\mathbf{p}_r\big\|^2$ and $\ \big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle\ $ to recover possible values for $\ \mathbf{x}_0\ $ and $\ \mathbf{v}\ $.
However, when $\ R=2\ $, and $\ U:\mathbb{R}^3\rightarrow\mathbb{R}^3\ $ is any rotation about the line through $\ \mathbf{p}_1\ $ and $\ \mathbf{p}_2\ $, then $\ U(\mathbf{x})=\mathbf{u}_0+\Theta\,\mathbf{x}\ $ for some orthogonal matrix $\ \Theta\ $ with $\ \mathbf{p}_r-\Theta\mathbf{p}_r=\mathbf{u}_0\ $ for $\ r=1,2\ $ $\big($because $\ U(\mathbf{p})=\mathbf{p}\ $ for all points $\ \mathbf{p}\ $ lying on the line through $\ \mathbf{p}_1\ $ and $\ \mathbf{p}_2\ \big)$, and
\begin{align}
\big\|\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big\|&=\big\|\Theta\big(\big(t_i-t_1\big)\mathbf{v}+\mathbf{x}_0-\mathbf{p}_r\big)\big\|\\
&=\big\|\big(t_i-t_1\big)\Theta\mathbf{v}+\Theta\mathbf{x}_0-\Theta\mathbf{p}_r\big\|\\
&=\big\|\big(t_i-t_1\big)\Theta\mathbf{v}+\Theta\mathbf{x}_0+\mathbf{u}_0-\mathbf{p}_r\big\|\ .
\end{align}
Thus, if $\ \mathbf{v}\ $,$\ \mathbf{x}_0\ $ and $\ t_i\ $, $\ i=1,2,\dots,T\ $, is any solution for the system of equations $(1)$, then so is $\ \mathbf{v}'=\Theta\,\mathbf{v}\ $, $\ \mathbf{x}_0'=\Theta\,\mathbf{x}_0+\mathbf{u}_0\ $ and $\ t_i\ $, $\ i=$$1,2,$$\dots,T\ $. Thus, with only two receivers it's impossible to recover $\ \mathbf{v}\ $ or $\ \mathbf{x}_0\ $ completely. Nevertheless, with a sufficient number of received pulses, the speed $\ \|\mathbf{v}\|\ $ of the receiver should still be uniquely determined.
When $\ R=3\ $, and $\ \mathbf{p}_1$, $\mathbf{p}_2$, $\mathbf{p}_3\ $ are not collinear, we will have known values $\ v=\|\mathbf{v}\|\ $,
\begin{align}
d_r&=\big\|\mathbf{x}_0-\mathbf{p}_r\big\|\ ,\ \text{and}\\
e_r&=\big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle
\end{align}
for $\ r=1,2,3\ $. Since $\ \mathbf{p}_3-\mathbf{p}_1\ $ and $\ \mathbf{p}_3-\mathbf{p}_2\ $ are linearly independent, we can solve the linear equations
\begin{align}
d_3^2-d_1^2-\big\|\mathbf{p}_3\big\|^2+\big\|\mathbf{p}_1\big\|^2&=2\big\langle\mathbf{p}_1-\mathbf{p}_3,\mathbf{x}_0\big\rangle\\
d_3^2-d_2^2-\big\|\mathbf{p}_3\big\|^2+\big\|\mathbf{p}_2\big\|^2&=2\big\langle\mathbf{p}_2-\mathbf{p}_3,\mathbf{x}_0\big\rangle
\end{align}
to obtain $\ \mathbf{x}_0=\mathbf{a}+\lambda\mathbf{u}\ $, where $\ \mathbf{u}\ $ is a unit vector perpendicular to $\ \mathbf{p}_3-\mathbf{p}_1\ $ and $\ \mathbf{p}_3-\mathbf{p}_2\ $, $\ \mathbf{a}\ $ a known point lying in the plane of $\ \mathbf{p}_1$, $\mathbf{p}_2$, and $\mathbf{p}_3\ $, and $\ \lambda\ $ a real parameter whose value must be either $\ \sqrt{d_1^2-\big\|\mathbf{a}-\mathbf{p}_1\big\|^2}\ $ or $\ {-}\sqrt{d_1^2-\big\|\mathbf{a}-\mathbf{p}_1\big\|^2}\ $. If $\ e_r\ $ is value recovered for $\ \big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle\ $ from the numerical solution of the system of equations $(2)$, then for each of the two possible values $\ \mathbf{a}\pm\Big(\sqrt{d_1^2-\big\|\mathbf{a}-\mathbf{p}_1\big\|^2}\Big)\mathbf{u}\ $ of $\ \mathbf{x}_0\ $, we can now solve the linear equations
$$
e_r=\big\langle\mathbf{v},\mathbf{x}_0-\mathbf{p}_r\big\rangle
$$
for $\ r=1,2,3\ $ to obtain the corresponding value of $\ \mathbf{v}\ $. The solution of either of these systems will always be the reflection of the solution of the other in the plane of $\ \mathbf{p}_1$, $\mathbf{p}_2$, and $\mathbf{p}_3\ $.