As Agol said in a finitely generated profinite group every subgroup of finite index is open. Therefore, the topology is unique and detremined by the algebra. This was first proved by Serre for pro-$p$ groups and eventually Nikolov and Segal proved it for any profinite groups.

Now, take $\mathbb{F}_p[[t]]$ formal power series over a field of $p$-elements and take their its abelian group. Then it is a metric pro-$p$ group which is the same as being countably based at $1$. On the other hand, take a vector space over $\mathbb{F}_p$ of a countable dimension and take its pro-$p$ complition. I am almost sure (so you might want to check the details) that in both cases you have a vector space of dimesnion $2^{\aleph_0}$ so the groups are isomorphic abstractly. But in the first case the topology is cuontably based at $1$ (and therefore in any points) while it is not countably based at $1$ in the second case.

You can read more about similar situation in Wilson's book on profinite groups on the chapter on free group.

As Agol said in a finitely generated profinite group every subgroup of finite index is open. Therefore, the topology is unique and detremined by the algebra. This was first proved by Serre for pro-$p$ groups and eventually Nikolov and Segal proved it for any profinite groups.

Now, take $\mathbb{F}_p[[t]]$ formal power series over a field of $p$-elements and take their its abelian group. Then it is a metric pro-$p$ group which is the same as being countably based at $1$. On the other hand, take a vector space over $\mathbb{F}_p$ of a countable dimension and take its pro-$p$ complition. I am almost sure (so you might want to check the details) that in both cases you have a vector space of dimesnion $2^{\aleph_0}$ so the groups are isomorphic abstractly. But in the first case the topology is countably based at $1$ (and therefore in any points) while it is not countably based at $1$ in the second case.

You can read more about similar situation in Wilson's book on profinite groups on the chapter on free group.