No. Consider, for example, the quantum complete intersection $A = k\langle X,Y\rangle/(X^2, Y^3, XY-qYX)$, $q\in k\setminus\{0\}$. This is a Frobenius local algebra (see Section 3 of arXiv:0709.3029), and the ideal $(X^2, Y^3, XY-qYX)$ in $k\langle X,Y\rangle$ is admissible.

Any isomorphism $f:A\to A^{\rm op}$ must satisfy $f(X) = aX+r$ and $f(Y) = bY + s$, where $a,b\in k\setminus\{0\}$ and $r,s\in\mathop{\rm rad}^2(A) = (XY, Y^2)$. Now $$f(XY) = f(Y)f(X) = (bY + s)(aX +r) = abq^{-1}XY + t,\quad t\in\mathop{\rm rad}\nolimits^3(A)$$ but also $$f(XY) = f(qYX) = qf(X)f(Y) = q(aX + r)(bY + s) = qabXY + qXu,\quad u\in\mathop{\rm rad}\nolimits^3(A)$$

Since $XY\notin\mathop{\rm rad}^3(A^{\rm op})$, this means that $abq^{-1}XY = qab XY$ and thus $q=q^{-1}$ (because $a,b\ne0$).

So unless $q = \pm1$, the algebras $A$ and $A^{\rm op}$ are not isomorphic.

No. Consider, for example, the quantum complete intersection $A = k\langle X,Y\rangle/(X^2, Y^3, XY-qYX)$, $q\in k\setminus\{0\}$. This is a Frobenius local algebra (see Section 3 of arXiv:0709.3029), and the ideal $(X^2, Y^3, XY-qYX)$ in $k\langle X,Y\rangle$ is admissible.

Any isomorphism $f:A\to A^{\rm op}$ must satisfy $f(X) = aX+r$ and $f(Y) = bY + s$, where $a,b\in k\setminus\{0\}$ and $r,s\in\mathop{\rm rad}^2(A) = (XY, Y^2)$. Now $$f(XY) = f(Y)f(X) = (bY + s)(aX +r) = abq^{-1}XY + t,\quad t\in\mathop{\rm rad}\nolimits^3(A)$$ but also $$f(XY) = f(qYX) = qf(X)f(Y) = q(aX + r)(bY + s) = qabXY + u,\quad u\in\mathop{\rm rad}\nolimits^3(A)$$

Since $XY\notin\mathop{\rm rad}^3(A^{\rm op})$, this means that $abq^{-1}XY = qab XY$ and thus $q=q^{-1}$ (because $a,b\ne0$).

So unless $q = \pm1$, the algebras $A$ and $A^{\rm op}$ are not isomorphic.

No. Consider, for example, the quantum complete intersection $A = k\langle X,Y\rangle/(X^2, Y^3, XY-qYX)$, $q\in k\setminus\{0\}$. This is a Frobenius local algebra (see Section 3 of arXiv:0709.3029), and the ideal $(X^2, Y^3, XY-qYX)$ in $k\langle X,Y\rangle$ is admissible.

Any isomorphism $f:A\to A^{\rm op}$ must satisfy $f(X) = aX$ and $f(Y) = bY + r$, where $a,b\in k\setminus\{0\}$ and $r\in\mathop{\rm rad}^2(A) = (XY, Y^2)$. Now $$f(XY) = f(Y)f(X) = (bY + r)aX = abq^{-1}XY + arX$$ but also $$f(XY) = f(qYX) = qf(X)f(Y) = qaX(bY + r) = qabXY + qXr$$

Since $rX,Xr\in\mathop{\rm rad}^3(A^{\rm op})\not\ni XY$, this means that $abq^{-1}XY = qab XY$ and thus $q=q^{-1}$ (because $a,b\ne0$).

So unless $q = \pm1$, the algebras $A$ and $A^{\rm op}$ are not isomorphic.

No. Consider, for example, the quantum complete intersection $A = k\langle X,Y\rangle/(X^2, Y^3, XY-qYX)$, $q\in k\setminus\{0\}$. This is a Frobenius local algebra (see Section 3 of arXiv:0709.3029), and the ideal $(X^2, Y^3, XY-qYX)$ in $k\langle X,Y\rangle$ is admissible.

Any isomorphism $f:A\to A^{\rm op}$ must satisfy $f(X) = aX+r$ and $f(Y) = bY + s$, where $a,b\in k\setminus\{0\}$ and $r,s\in\mathop{\rm rad}^2(A) = (XY, Y^2)$. Now $$f(XY) = f(Y)f(X) = (bY + s)(aX +r) = abq^{-1}XY + t,\quad t\in\mathop{\rm rad}\nolimits^3(A)$$ but also $$f(XY) = f(qYX) = qf(X)f(Y) = q(aX + r)(bY + s) = qabXY + qXu,\quad u\in\mathop{\rm rad}\nolimits^3(A)$$

Since $XY\notin\mathop{\rm rad}^3(A^{\rm op})$, this means that $abq^{-1}XY = qab XY$ and thus $q=q^{-1}$ (because $a,b\ne0$).

So unless $q = \pm1$, the algebras $A$ and $A^{\rm op}$ are not isomorphic.