Fix an algebraically closed field $F$. Are there only finitely many symmetric algebras with unit over $F$ of a given finite dimension (up to isomorphism)? By symmetric I mean a Frobenius algebra where the form is symmetric. For a general associative algebra I know it is false:
Are there only finitely many associative algebras of fixed dimension?
However, the example in the above thread does not give an infinite number of non-isomorphic symmetric algebras.
The answer is certainly "no", but I can't immediately think of an example where it's easy to prove all the details.
In Karin Erdmann's (almost) determination of the structure of tame blocks, there are one-parameter families of algebras, where it's not known which parameters give algebras that actually occur as blocks of finite group algebras. I'm pretty sure those will give a counterexample.
Alternatively, "trivial extensions" give a construction of symmetric algebras. If $A$ is any finite-dimensional algebra, then $TA=A\oplus DA$, where $DA$ is the vector space dual of $A$ considered as an $A$-bimodule in the obvious way, and with multiplication $$(a,\theta)(b,\phi)=(ab,a\phi+\theta b)$$ is symmetric, and although non-isomorphic algebras may have isomorphic trivial extensions, I'm sure that doesn't typically happen, and that you could could construct infinite families of non-isomorphic trivial extension algebras starting from infinite families of non-isomorphic non-symmetric algebras.
Edit: Actually, here's a fairly simple example in the same spirit as my answer to the question you linked to.
For $0\neq a\in F$, let $$S(a)=F\langle x,y,z\vert x^2=y^2=z^2=0,xy=ayx,yz=azy,zx=axz\rangle.$$ So $S(a)$ is an $8$-dimensional algebra with basis $\{1,x,y,z,xy,yz,zx,xyz\}$, and $\operatorname{rad}^iS(a)$ is spanned by the basis elements of degree at least $i$ in $\{x,y,z\}$.
It is symmetric, with a symmetrizing form that kills all the basis elements except $xyz$.
Unless $a=-1$, the elements $w$ of $\operatorname{rad}S(a)/\operatorname{rad}^2S(a)$ such that $w^2=0$ in $\operatorname{rad}^2S(a)/\operatorname{rad}^3S(a)$ are precisely the scalar multiples of $x$, $y$ and $z$. Choosing two linearly independent such elements $w_1,w_2$, either $w_1w_2=aw_2w_1$ or $w_2w_1=aw_1w_2$ in $\operatorname{rad}^2S(a)/\operatorname{rad}^3S(a)$. So $S(a)\not\cong S(b)$ unless $a=b$ or $a=b^{-1}$.
