If I have a power series
$$\sum_{j = 1}^{\infty}a_jx^{2j+1} = x\sum_{j = 1}^{\infty}a_jx^{2j} $$
Given that I have the radius of convergence $R$ of
$$\sum_{j = 1}^{\infty}a_jz^{j}$$
where $z = x^2$
What can I say about the original series?
This may be clear that they have the same radii of convergence, but not to me.
No, the series in $z$ and the ones in $x$ do not have the same radius of convergence. To see this, there are two approaches.
The first has already been sketched in your question and the subsequent comments. Calling $z = x^2$, if the series $\sum a_j z^j$ has the radius $R$, then it converges for $|z| < R$, which means $|x^2| < R$ or, equivalently, $|x| < \sqrt R$ (this is true because $x \sum a_j x^{2j}$ and $\sum a_j x^{2j}$ have the same radius). This approach is very clear and simple, but it fails if you modify your original question a bit: what about $\sum a_j x^{j!}$? You may always write $x^{2j+1}$ as $x z^j$, but how do you proceed with $x^{j!}$?
A second approach, that works in general, is based upon noting that your series "skips" some powers, so let us put them back by multiplying them by $0$. Your series can thus be rewritten as $\sum b_j x^j$ where $b_j = \left\{ \begin{matrix} a_{\frac {j-1} 2}, & 2 \nmid j \\ 0, & 2 \mid j \end{matrix} \right.$. How do we compute the radius of convergence, now? Usually, students are taught the formula $R = \dfrac 1 {\lim \sqrt[j] {|b_j|}}$, but it seems that in our case the limit in the denominator might not exist (it seems that there are at least two limit points, one of them being $0$, corresponding to the subsequence formed by $b_2 = b_4 = b_6 = \dots = 0$). The answer is to use a more complicated but also more general formula: $R = \dfrac 1 {\limsup \sqrt[j] {|b_j|}}$. For simplicity, think of $\limsup$ as being the largest of all the limit points, so
$$\limsup \limits _j \sqrt[j] {|b_j|} = \max \left( \limsup \limits _{j \ \text{even}} \sqrt[j] {| b_j |}, \limsup \limits _{j \ \text{odd}} \sqrt[j] {| b_j |} \right) = \max \left( 0, \limsup \limits _{j \ \text{odd}} \sqrt[j] {| b_j |} \right) = \limsup \limits _{j \ \text{odd}} \sqrt[j] {| b_j |} = \limsup \limits _{j \ \text{odd}} \sqrt[j] {\left| a_{\frac {j-1} 2} \right|} = \limsup \limits _{j = 2k + 1} \sqrt[j] {\left| a_{\frac {j-1} 2} \right|} = \limsup \limits _k \sqrt[2k + 1] {| a_k |} = \limsup \limits _k \sqrt[2k] {| a_k |} = \sqrt {\limsup \limits _k \sqrt[k] {| a_k |}} .$$
This shows that if $R$ is the convergence radius of $\sum a_k z^k$ and $R'$ is the convergence radius of $\sum a_j x^{2j+1} = \sum b_j x^j$, then $R' = \sqrt R$. As said above, this approach is slightly more technical, but manages to solve a larger class of problems that the first approach.
