Can a (finite dimenaional) $\mathbb{K}$-algebra $A$ be equipped with more than one Frobenius structure $\lambda:A \to \mathbb{K}$? Of course we identify two structures $\lambda$ and $\lambda'$ if they differ by a scalar multiple.
If it can what is a good example? If we restrict to filtered Frobenius algebras can this help with uniqueness?
If $A$ is a Frobenius $K$-algebra and $\lambda\colon A\to K$ is a Frobenius form, then the Frobenius forms are the mappings of the form $a\mapsto \lambda(ua)$ with $u$ a unit of $A$.
One way to see that is being Frobenius means $A_A\cong \hom_K({}_AA,K)$ (where $M_A$ means $M$ is a right $A$-module and ${}_AM$ means $M$ is a left $A$-module). A Frobenius form is precisely the image of $1$ under such an isomorphism. Since the group of units of $A$, acting via left multiplication, is the automorphism group of $A$ we see that any isomorphism takes $1$ to $\lambda u$ for some unit $u$.
For example, in @JaSch's answer $\lambda_2=\lambda_1(-i(a+ib))$ so the unit $u$ is $-i$.
I think the following should give an example. Let $ \mathbb{K} = \mathbb{R} $ and $ A = \mathbb{C} $. Moreover, we define $ \lambda_1, \lambda_2 \colon \mathbb{C} \to \mathbb{R} $ by $$ \lambda_1 \left( a+ib \right) = a,\quad \lambda_2 \left( a+ib \right) = b. $$ Clearly, these maps are $ \mathbb{R} $-linear and do not differ by a scalar multiple. Furthermore, neither the kernel of $ \lambda_1 $ nor the kernel of $ \lambda_2 $ can contain a non-zero ideal of $ A = \mathbb{C} $ because it is a field.
