There exist no ring homomorphism $\sigma$ between $\mathbb{Q}_3$ and $\mathbb{Q}_5$ with $\sigma(1)=1$.
Approach: Since $\sigma(1)=1$, we can conclude that $\sigma(q) = q$ for all $q \in \mathbb{Q}$. We have that $$1+3+3^2+3^3+... \to (1-3)^{-1}$$ in $\mathbb{Q}_3$. But this does not hold in $\mathbb{Q}_5$. Therefore there exist no such ring homomorphism. Is that correct?
On
No. The ring homomorphism need not be continuous with respect to the $3$- and $5$-adic (or any) topology.
Instead, look at the polynomial in $X^2 - 10$ in both fields. Use that $ℚ_5$ is the quotient field of the factorial ring $ℤ_5$ (in which $5$ is prime) and use Hensel on the polynomial in $ℚ_3$.
(Note that $ℚ_3$ is a field, so any such ring homomorphism $σ$ would be an embedding.)
(First, no, there is assumption of continuity, so convergence arguments are not directly relevant...)
Thinking in terms of Hensel's lemma, there will be different square roots (of rational integers) in the two $p$-adic fields. For example, exactly the rational integers $n=\pm 1\mod 5$ have square roots in $\mathbb Z/5$, and then, by Hensel's lemma, in $\mathbb Z_5\subset \mathbb Q_5$. Similarly, exactly the rational integers $n=1\mod 3$. (One should also note that a square root $\alpha$ in $\mathbb Z_p$ of a rational integer $n$ is integral over $\mathbb Z$, so is integral over $\mathbb Z_p$, so is in $\mathbb Z_p$ rather than $\mathbb Q_p$.)
Thus, for example, there is a $\sqrt{7}$ in $\mathbb Q_3$, but not in $\mathbb Q_5$. Oppositely, there is a $\sqrt{-1}$ in $\mathbb Q_5$, but not in $\mathbb Q_3$. So there can be no inclusion of either in the other (regardless of topologies).