Building on G.Myerson's answer and KConrad's explanation, it's not hard to construct pairs $K,K'$ of quartic fields that have both the same discriminant and the same regulator.

Namely, start with a real quadratic field $F\phantom.$ with fundamental unit $\epsilon$, and let $K,K'$ be totally imaginary quadratic extensions of $F$, not isomorphic with $F\phantom.((-\epsilon)^{1/2})$ and with no roots of unity other than $\pm 1$, whose relative discriminants $d_{K/F}$ and $d_{K'/F}$ have the same norm in ${\bf Q}$. Then $K$ and $K'$ have the same discriminant over ${\bf Q}$, and each has the same unit group $\pm \epsilon^{\bf Z}$ as $F$, so they have the same regulator.

For an explicit example, take $F = {\bf Q}(r)$ with $r=\sqrt{2}$, and let $K = F\phantom.(\sqrt{ab})$ and $K'=F\phantom.(\sqrt{a'b})$ where $a,a' = 7 \pm 2r$ have norm $41$ and $b=5+2r$ has norm $17$. Then $K$ and $K'$ are generated by roots of $x^4+54x^2+697$ and $x^4+86x^2+697$, and are not isomorphic (e.g. the rational prime $7$ splits completely in $K'$ but not in $K$) but both have discriminant $44608 = 2^6 17 \cdot 41$ and unit group $\pm \epsilon^{\bf Z}$ where $\epsilon=1+r$.

The same technique generates arbitrarily large packets of quartic fields with the same discriminant and regulator. More generally, for any totally real field $F\phantom.$ of degree $d>1$ there are arbitrarily large packets of totally imaginary quadratic extensions $K$ of $F\phantom.$ with the same discriminant over ${\bf Q}$ and the same unit group as $F\phantom.$: by the Dirichlet unit theorem each $K$ has the same unit rank as $F$, so — as long as $K$ has no new roots of unity and is not generated by the square root of a unit of $F\phantom.$ — all the units of $K$ are contained in $F$. [The requirement that $K$ have no roots of unity other than $\pm 1$ is used for this conclusion $O_K^* = O_F^*\phantom.$, but is not needed for equality of regulators.]

Degree $4$ is likely minimal here: in degree $2$ (and $1$), number fields are uniquely determined by their discriminant; and as for degree $3$, while there can be arbitrarily large packets of cubic number fields of the same discriminant, it seems most unlikely (though hard to disprove in the totally real case) that any two would have the same regulator.

Building on G.Myerson's answer and KConrad's explanation, it's not hard to construct pairs $K,K'$ of quartic fields that have both the same discriminant and the same regulator. [Edited to add examples where $K$ and $K'$ do not have the same roots of unity.]

Namely, start with a real quadratic field $F\phantom.$ with fundamental unit $\epsilon$, and let $K,K'$ be totally imaginary quadratic extensions of $F$, not isomorphic with $F\phantom.((-\epsilon)^{1/2})$ and with no roots of unity other than $\pm 1$, whose relative discriminants $d_{K/F}$ and $d_{K'/F}$ have the same norm in ${\bf Q}$. Then $K$ and $K'$ have the same discriminant over ${\bf Q}$, and each has the same unit group $\pm \epsilon^{\bf Z}$ as $F$, so they have the same regulator.

For an explicit example, take $F = {\bf Q}(r)$ with $r=\sqrt{2}$, and let $K = F\phantom.(\sqrt{ab})$ and $K'=F\phantom.(\sqrt{a'b})$ where $a,a' = 7 \pm 2r$ have norm $41$ and $b=5+2r$ has norm $17$. Then $K$ and $K'$ are generated by roots of $x^4+54x^2+697$ and $x^4+86x^2+697$, and are not isomorphic (e.g. the rational prime $7$ splits completely in $K'$ but not in $K$) but both have discriminant $44608 = 2^6 17 \cdot 41$ and unit group $\pm \epsilon^{\bf Z}$ where $\epsilon=1+r$.

The same technique generates arbitrarily large packets of quartic fields with the same discriminant and regulator. More generally, for any totally real field $F\phantom.$ of degree $d>1$ there are arbitrarily large packets of totally imaginary quadratic extensions $K$ of $F\phantom.$ with the same discriminant over ${\bf Q}$ and the same unit group as $F\phantom.$: by the Dirichlet unit theorem each $K$ has the same unit rank as $F$, so — as long as $K$ has no new roots of unity and is not generated by the square root of a unit of $F\phantom.$ — all the units of $K$ are contained in $F$.

[The requirement that $K$ have no roots of unity other than $\pm 1$ is used for this conclusion $O_K^* = O_F^*\phantom.$, but is not needed for equality of regulators. EDIT Indeed it may happen that in such a pair of quartic fields $K$ had more roots of unity than $K'$: if $\epsilon \equiv 1 \bmod 4$ then ($\epsilon$ is totally positive and) $K=F\phantom.(\sqrt{-3})$ has sixth roots of unity while $K'=F\phantom.(\sqrt{-3\epsilon})$ does not. The regulators are still the same unless $K = F\phantom.((-\epsilon)^{1/2})$, that is, unless $3\epsilon$ is a square in $F$, in which case the regulator of $K'$ is twice that of $K$. For example we can take $F = {\bf Q}(\sqrt{203})$, which has $\epsilon = 57 + 4 \sqrt{203}$, but not $F = {\bf Q}(\sqrt{39})$ because then $\epsilon = 25 + 4 \sqrt{39} = (6+\sqrt{39})^2/3$ so $K$ contains the square roots of $-\epsilon$. TIDE]

Degree $4$ is likely minimal here: in degree $2$ (and $1$), number fields are uniquely determined by their discriminant; and as for degree $3$, while there can be arbitrarily large packets of cubic number fields of the same discriminant, it seems most unlikely (though hard to disprove in the totally real case) that any two would have the same regulator.

Building on G.Myerson's answer and KConrad's explanation, it's not hard to construct pairs $K,K'$ of quartic fields that have both the same discriminant and the same regulator.

Namely, start with a real quadratic field $F\phantom.$ with fundamental unit $\epsilon$, and let $K,K'$ be totally imaginary quadratic extensions of $F$, not isomorphic with $F\phantom.(-\epsilon^{1/2})$ and with no roots of unity other than $\pm 1$, whose relative discriminants $d_{K/F}$ and $d_{K'/F}$ have the same norm in ${\bf Q}$. Then $K$ and $K'$ have the same discriminant over ${\bf Q}$, and each has the same unit group $\pm \epsilon^{\bf Z}$ as $F$, so they have the same regulator. For an explicit example, take $F = {\bf Q}(r)$ with $r=\sqrt{2}$, and let $K = F\phantom.(\sqrt{ab})$ and $K'=F\phantom.(\sqrt{a'b})$ where $a,a' = 7 \pm 2r$ have norm $41$ and $b=5+2r$ has norm $17$. Then $K$ and $K'$ are generated by roots of $x^4+54x^2+697$ and $x^4+86x^2+697$, and are not isomorphic (e.g. the rational prime $7$ splits completely in $K'$ but not in $K$) but both have discriminant $44608 = 2^6 17 \cdot 41$ and unit group $\pm \epsilon^{\bf Z}$ where $\epsilon=1+r$.

The same technique generates arbitrarily large packets of quartic fields with the same discriminant and regulator. More generally, for any totally real field $F\phantom.$ of degree $d>1$ there are arbitrarily large packets of totally imaginary quadratic extensions $K$ of $F\phantom.$ with the same discriminant over ${\bf Q}$ and the same unit group as $F\phantom.$: by the Dirichlet unit theorem each $K$ has the same unit rank as $F$, so — as long as $K$ has no new roots of unity and is not generated by the square root of a unit of $F\phantom.$ — all the units of $K$ are contained in $F$.

Degree $4$ is likely minimal here: in degree $2$ (and $1$), number fields are uniquely determined by their discriminant; and while there can be arbitrarily large packets of cubic number fields of the same discriminant, it seems most unlikely (though hard to disprove in the totally real case) that any two would have the same regulator.

Building on G.Myerson's answer and KConrad's explanation, it's not hard to construct pairs $K,K'$ of quartic fields that have both the same discriminant and the same regulator.

Namely, start with a real quadratic field $F\phantom.$ with fundamental unit $\epsilon$, and let $K,K'$ be totally imaginary quadratic extensions of $F$, not isomorphic with $F\phantom.((-\epsilon)^{1/2})$ and with no roots of unity other than $\pm 1$, whose relative discriminants $d_{K/F}$ and $d_{K'/F}$ have the same norm in ${\bf Q}$. Then $K$ and $K'$ have the same discriminant over ${\bf Q}$, and each has the same unit group $\pm \epsilon^{\bf Z}$ as $F$, so they have the same regulator.

For an explicit example, take $F = {\bf Q}(r)$ with $r=\sqrt{2}$, and let $K = F\phantom.(\sqrt{ab})$ and $K'=F\phantom.(\sqrt{a'b})$ where $a,a' = 7 \pm 2r$ have norm $41$ and $b=5+2r$ has norm $17$. Then $K$ and $K'$ are generated by roots of $x^4+54x^2+697$ and $x^4+86x^2+697$, and are not isomorphic (e.g. the rational prime $7$ splits completely in $K'$ but not in $K$) but both have discriminant $44608 = 2^6 17 \cdot 41$ and unit group $\pm \epsilon^{\bf Z}$ where $\epsilon=1+r$.

The same technique generates arbitrarily large packets of quartic fields with the same discriminant and regulator. More generally, for any totally real field $F\phantom.$ of degree $d>1$ there are arbitrarily large packets of totally imaginary quadratic extensions $K$ of $F\phantom.$ with the same discriminant over ${\bf Q}$ and the same unit group as $F\phantom.$: by the Dirichlet unit theorem each $K$ has the same unit rank as $F$, so — as long as $K$ has no new roots of unity and is not generated by the square root of a unit of $F\phantom.$ — all the units of $K$ are contained in $F$. [The requirement that $K$ have no roots of unity other than $\pm 1$ is used for this conclusion $O_K^* = O_F^*\phantom.$, but is not needed for equality of regulators.]

Degree $4$ is likely minimal here: in degree $2$ (and $1$), number fields are uniquely determined by their discriminant; and as for degree $3$, while there can be arbitrarily large packets of cubic number fields of the same discriminant, it seems most unlikely (though hard to disprove in the totally real case) that any two would have the same regulator.