In dimension $n\leq 3$, "isotropic" is synonymous with "rotationally symmetric", and the latter is only slightly stronger in dimensions $n>3$. In any dimension $n>1$, there are exactly four connected Riemannian manifolds which are rotationally symmetric. These are $\mathbb{R}^n$, hyperbolic space $\mathbb{H}^n$, the hypersphere $S^n$, and the Real projective space $\mathbb{RP}^n$. This answers your 5th question.
We say a manifold is rotationally symmetric if for any point $p$ and any orthonormal basis $v_1,\cdots,v_n\in T_p(M)$, and any other orthonormal basis $w_1,\cdots,w_n\in T_p(M)$, there's an isometry $f$ such that generally $f_*(v_k)=w_k$. We actually only require a weaker notion I'll call sectional symmetry: we can find an isometry $f$ such that $\operatorname{span}\{f_*(v_1),f_*(v_2)\}=\operatorname{span}\{w_1,w_2\}$. This is a mild strengthening of isotropy, which permits us to superimpose two-dimensional tangent planes in addition to how isotropy permits us to superimpose tangent lines. In dimension $n=3$, isotropy is equivalent to sectional symmetry, since superimposing the normal vectors of two planes results in superimposing the planes themselves.
Let $M$ be our manifold, which is isotropic and sectionally symmetric. As pointed out by Jason DeVito, every isotropic manifold is automatically homogeneous. By appealing to homogeneity and sectional symmetry, it follows that our manifold has constant sectional curvature, and moreover homogeneous manifolds are complete. Since we include that our manifold is connected, then it is a Space Form and subject to the Killing-Hopf theorem. That theorem asserts that our space is isometric (up to rescaling) to a quotient of one of $\mathbb{R}^n$, $\mathbb{H}^n$, or $S^n$. More specifically, that quotient is obtained by a group of isometries acting freely and properly discontinuously. The trivial quotients give $\mathbb{R}^n$, $\mathbb{H}^n$, and $S^n$, and the only nontrivial isotropic quotient gives $\mathbb{RP}^n$.
To prove my claim, let $\mathbb{M}$ denote the parent manifold, which must be (a rescaling of) one of $\mathbb{R}^n$, $\mathbb{H}^n$, or $S^n$. Let $G$ denote our group of isometries, and let $\sim$ denote the equivalence relation on $\mathbb{M}$ induced by $G$, so $M\cong(\mathbb{M}/\sim)$ is our isotropic manifold. We assume $G$ is nontrivial, meaning $\sim$ is not the identity relation. Given any point $x\in \mathbb{M}$, and any bounded neighborhood $U$ of $x\in U\subseteq\mathbb{M}$, it must hold that there are only finitely many $y\in U$ with $x\sim y$. This is a consequence of the fact that $G$ acts properly discontinuously.
Since $\sim$ is nontrivial, there exists a point $y\neq x$ having $y\sim x$. Moreover since $\sim$ is locally discrete in the sense just established, we can select $y$ to minimize the distance $d(x,y)$. Let $\gamma:[0,1]\to\mathbb{M}$ be a minimal-length geodesic obeying $\gamma(0)=x$ and $\gamma(1)=y$. Let $\tilde{\gamma}:[0,1]\to M$ denote the image of $\gamma$ under the $\sim$ quotient, then $\tilde{\gamma}(0)=\tilde{\gamma}(1)$. Let $\ell$ denote the length of $\gamma$, which is nonzero since $x\neq y$. The length of $\tilde{\gamma}$ must also be $\ell$, due to the minimality properties imposed on $y$ and on $\gamma$. By consequence of isotropy and homogeneity of $M$, every geodesic in $M$ of length $\ell$ must also be a closed loop, just like $\tilde{\gamma}$.
Let $z\in\mathbb{M}$ be any point such that $d(x,z)=d(x,y)$. Let $\sigma:[0,1]\to \mathbb{M}$ denote a minimal-length geodesic from $x$ to $z$, which must also have length $\ell$ due to the isotropy of $\mathbb{M}$. Let $\tilde{\sigma}$ denote the image of $\sigma$ under our quotient, which must have length $\ell$ due to the minimality imposed on $\sigma$ and on $y$. It follows that $\tilde{\sigma}(0)=\tilde{\sigma}(1)$, and therefore $x\sim z$. Letting $U=\{z\in \mathbb{M} : d(x,z)\leq d(x,y)\}$, recall that since $U$ is bounded, there can only be finitely many $z\in U$ for which $z\sim x$. Therefore, there can only be finitely many $z$ for which $d(x,z)=d(x,y)$. This is impossible in Euclidean space. It's also impossible in Hyperbolic space. It necessarily follows that $\mathbb{M}=S^n$ is the sphere, and moreover $y=-x$ must be the antipodal point of $x$. Therefore $\sim$ is the antipodal relation, and $M \cong (\mathbb{M}/\sim) = \mathbb{RP}^n$.
The above proof doesn't work in dimension $n=1$, since zero-dimensional circles are finite, but the characterization in that case is trivial: $\mathbb{H}^1 = \mathbb{R}^1$, and $S^1$ is the only nontrivial quotient of $\mathbb{R}$. As a minor nitpick, I'd also like to point out that the term "isotropic" in physics usually means "isotropic from our perspective", which is different from having isotropy from all perspectives. The latter implies homogeneity as established, but the former does not.
