Solve Group Equation Given Two Equations Given the group equations $x^{2}=a^{2}$ and $x^{5}=e$, I attempted to solve for $a$ by substituting $a^{2}x^{3}$ into $x^{5}=e$.

Then I solved for $x$ by noting $a^{2}x^{3} = e \implies (a^{2}x^{3})x^{-3} = (e)x^{-3}\implies a^{2}=x^{-3}$.

Is this correct? If so, can it be broken down further? Why/why not?
If you are trying to find the structure of the group, you might proceed as follows:

$$a^{10}=(x^2)^5=(x^5)^2=e$$

You have $x^2=a^2$ whence $x^4=a^4$ and $x^6=a^6$ but also $x^6=x^5x=x$ so that $x=a^6$, and $x^3=a^8$.

You can check that if $a^{10}=e$ and $x=a^6$ the cyclic group of order $10$ generated by $a$ satisfies the given relations. No larger group can do so, but a smaller group might - for example the trivial group with $a=x=e$. I'll leave you to explore the other cases which may arise.

Note: within the cyclic group of order $10$ the powers of $x$ are all even powers of $a$, so you don't get a nice equation $a=x^n$ with $n$ an integer.
We have, $x^2= a^2$ and $x^5 = e$
Now,

$X^5= e$

or, $x^2. x^2. x= e$

or, $a^2 . a^2 . x =e$ ( substituting from 1st equation)

or, $a^4 . x= e$

Operating both sides by $a^{-4}$ on the left. We get,

$(a^{-4}. a^4). x= a^{-4}. e$

or, $e. x = a^{-4}$
Therefore, $x= a^{-4}$